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This article is cited in 2 scientific papers (total in 2 papers) On Jutila's integral in the circle problem M. A. Koroleva, D. A. Popovb a Steklov Mathematical Institute of Russian Academy of Sciences, Moscow b Lomonosov Moscow State University, Belozersky Research Institute of Physico-Chemical Biology
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2022, Volume 86, Issue 3, DOI: https://doi.org/10.4213/im9155 
Bibliographic databases:
    To the blessed memory of Professor Aleksandar Ivić (6.3.1949–27.12.2020) § 1. IntroductionGiven an arbitrary $x\geqslant 1$, we write $\Delta(x)$ for the difference where $\tau(n)$ is the number of divisors of an integer $n\geqslant 1$ and $\gamma$ is the Euler constant. Besides the classical Dirichlet divisor problem, which consists in proving the bound $\Delta(x) = O(x^{1/4+\varepsilon})$ ($x\to +\infty$, $\varepsilon>0$ is arbitrarily small and fixed), there are other unsolved problems related to the behaviour of $\Delta(x)$. One of them is Jutila’s conjecture1 saying that the following bound holds as $x\to +\infty$ for every $U$, $11\leqslant U\ll \sqrt{x}$: Proving (1) seems to be a very difficult problem. For example, it is known that (1) yields a solution of the divisor problem.To characterize the behaviour of $\Delta(x+U)-\Delta(x)$ ‘in the mean’, Jutila [2] introduced the integral
$$
\begin{equation}
\mathcal{Q}_{\Delta}=\mathcal{Q}_{\Delta}(X;H,U) =\int_{X}^{X+H}\bigl(\Delta(x+U)-\Delta(x)\bigr)^2\, dx
\end{equation}
\tag{2}
$$
and obtained an explicit formula for it in the case when the parameters satisfy the condition $1\leqslant U\ll \sqrt{X}\ll H\leqslant X$. A corollary of this formula is the following double bound, which is of the correct order of magnitude:
$$
\begin{equation}
\mathcal{Q}_{\Delta}\asymp HU\biggl(\log{\frac{\sqrt{X}}{U}}\biggr)^3,
\end{equation}
\tag{3}
$$
provided that, in addition, $HU\gg X^{1+\varepsilon}$. In particular, it follows from (3) that Jutila’s conjecture holds ‘for almost all’ $x$. See [1] and [3] for the relation of Jutila’s results to the circle problem. In this paper we consider an analogue of $\mathcal{Q}_{\Delta}$ for the function $P(t)$, that is, the quantity
$$
\begin{equation*}
\mathcal{Q}_{P}=\mathcal{Q}_{P}(T;H,U)=\int_{T}^{T+H}\bigl(P(t+U)-P(t)\bigr)^2\, dt.
\end{equation*}
\notag
$$
Here $P(t)$ is the error term in the circle problem, that is, the difference where $r(n)$ is the number of representations of $n$ as the sum of two squares. Since the results in [2] related to $\Delta(x)$ are based on the so-called truncated Hardy–Voronoï formula,2 they can be transferred to $P(t)$ with obvious changes. Conversely, all our results about $P(t)$ can be transferred to the case of $\Delta(x)$. Here is Jutila’s theorem for $P(t)$. Jutila’s theorem. Let $T\to +\infty$ and suppose that $1\leqslant U \ll \sqrt{T}\ll H\leqslant T$. Then The cube of the logarithm in (3) is replaced by the logarithm in (5) because of the different behaviour of the mean values of $\tau^2(n)$ and $r^2(n)$, namely,
$$
\begin{equation*}
\sum_{n\leqslant x}\tau^2(n)\sim\frac{1}{\pi}x(\log{x})^3,\qquad \sum_{n\leqslant t}r^2(n)\sim 4t\log{t}.
\end{equation*}
\notag
$$
We introduce the ‘correlation function’
$$
\begin{equation*}
\mathcal{K}_{P}=\mathcal{K}_{P}(T;H,U)=\int_{T}^{T+H}P(t+U)P(t)\, dt.
\end{equation*}
\notag
$$
We can easily conclude from the definition of $\mathcal{Q}_{P}$ that
$$
\begin{equation}
2\mathcal{K}_{P}(T;H,U)=I(T+U,H)+I(T,H)-\mathcal{Q}_{P}(T;H,U),
\end{equation}
\tag{6}
$$
where is the second moment of the function $P(t)$ on the interval $T\leqslant t\leqslant T+H$. The asymptotic behaviour of the second moment of $P(t)$ on the interval $0\leqslant t\leqslant T$ was described by Landau:
$$
\begin{equation}
\begin{gathered} \, \int_0^{T}P^2(t)\,dt=\frac{A}{3\pi^2}T^{3/2}+O(T^{1+\varepsilon}), \\ A=\sum_{n=1}^{+\infty}\frac{r^2(n)}{n^{3/2}} =\frac{16\zeta^2(3/2)L^2(3/2,\chi_4)}{\zeta(3)(1+2^{-3/2})} =50.1560561426\dots \end{gathered}
\end{equation}
\tag{8}
$$
(the bound for the error term in (8) was later improved; see [4] for details). Here is an analogue of (8) for $I(T,H)$ in the case when $\sqrt{T}(\ln{T})^2=o(H)$ and $H = o(T)$: (see our Lemma 9). It follows from (4), (6) and (9) that the following asymptotic formula for the correlation function holds in the case when $U\geqslant T^{2\varepsilon}$, $U = o(\sqrt{T})$ and $T^{1+\varepsilon}U^{-1}\leqslant H = o(T)$:
$$
\begin{equation}
\mathcal{K}_{P}(T;H,U)=\frac{A}{2\pi^2}H\sqrt{T}\,\bigl(1+o(1)\bigr).
\end{equation}
\tag{10}
$$
Defining a quantity $k_{P}(T;H,U)$ by the equality
$$
\begin{equation*}
k_{P}(T;H,U)=\frac{2\pi^2}{H\sqrt{T}}\mathcal{K}_{P}(T;H,U),
\end{equation*}
\notag
$$
we can write (10) in the form At the same time, applying Cauchy’s inequality and the asymptotic formula (9) to the integral $\mathcal{K}_{P}(T;H,U)$, we conclude that
$$
\begin{equation}
\begin{gathered} \, |k_{P}(T;H,U)|\leqslant \frac{2\pi^2}{H\sqrt{T}}\sqrt{I(T,H)}\,\sqrt{I(T+U,H)}=A+o(1), \\ -A+o(1)\leqslant k_{P}(T;H,U)\leqslant A+o(1) \end{gathered}
\end{equation}
\tag{12}
$$
for all $T$, $H$ and $U$ such that $H, U = o(T)$ and $\sqrt{T}(\ln{T})^2 = o(H)$. Thus, (10) and (12) mean that, for any small $U$, $T^{\varepsilon}\ll U = o(\sqrt{T})$, and sufficiently large $H$, the correlation function attains the largest possible value. This agrees completely with the ‘naive’ interpretation of $\mathcal{K}_{P}$ as the measure of independence of $P(t)$ and $P(t+U)$ (regarded as random variables): if $U$ is small, then $P(t)$ and $P(t+U)$ are ‘strongly’ dependent and their correlation is large.3 It was mentioned in [2] that when $U\gg \sqrt{X}$, the quantities $\Delta(x+U)$ and $\Delta(x)$ must behave as independent random variables and, therefore, investigating $\mathcal{Q}_{\Delta}(X,H;U)$ is of no interest. A similar conclusion might be expected for the functions $\mathcal{Q}_{P}$ and $\mathcal{K}_{P}$. However, an analysis of the behaviour of $\mathcal{Q}_{P}$ and $\mathcal{K}_{P}$ in the case of large $U$, $U\gg \sqrt{T}$, reveals that their behaviour is more complicated than might be expected. In particular, besides the ‘barrier’ of order $\sqrt{T}$ for $U$, whose attainment drastically changes the behaviour of $\mathcal{Q}_{P}$, the ratio $\varkappa = HU/T^{3/2}$ begins to play an important role. When it is bounded above by a positive constant, the correlation function $\mathcal{K}_{P}$ does not attain its maximal positive value (Theorem 1). This corresponds to the intuitively expected effect of ‘relaxing’ the dependence between $P(t)$ and $P(t+U)$. In the case of ‘medium values’ of $\varkappa$, there are pairs $H, U$ with $\mathcal{K}_{P}$ maximally close to the upper bound, and there are pairs $H, U$ with $\mathcal{K}_{P}$ maximally close to the expected unimprovable lower bound (Theorem 2). Finally, when $\varkappa$ is very small, we establish unimprovable upper and lower bounds for $\mathcal{K}_{P}$ (of which the latter is different from the bound in (10)) and prove that, for any given $T$ and $H$, the quantity $\mathcal{K}_{P}$ is close to these upper and lower bounds on sets of values of $U$ of positive measure and, at the same time, the modulus of $\mathcal{K}_{P}$ is close to zero on sets of values of $U$ of positive measure (Theorems 3–5). In the case of small $\varkappa$, we are also able to prove the existence of pairs $H, U$ such that the correlation function takes values of anomalously small modulus (Theorem 6). We now proceed to give exact statements of our main results. Theorem 1. Let $\varepsilon_0>0$ be an arbitrary constant, $T\to +\infty$, $H,U = o(T)$, $\sqrt{T}(\log T)^2 = o(H)$, $\sqrt{T} = o(U)$ and suppose that $\varkappa = HU/T^{3/2}\geqslant \varepsilon_0$ when $T\,{\geqslant}\, T_0(\varepsilon_0)$. Moreover, assume that Then the following inequality holds: where $\delta$ can be chosen as $A - 4B/(\pi\varepsilon_0)$ when $\varepsilon_0\geqslant 1$ and as $A(\pi\varepsilon_0)^2/128$ when $0<\varepsilon_0<1$. Theorem 2. Let $\varepsilon_0, \varepsilon_1$ be sufficiently small absolute constants and let $D = D(T)$ be an unbounded function of arbitrarily slow growth as $T\to+\infty$ such that Theorem 3. Let $\varepsilon$ be arbitrarily small and fixed and let $D = D(T)$ be an unbounded function of arbitrarily slow growth as $T\to+\infty$. Assume that $H, V = o(T)$, $\sqrt{T} = o(V)$, $\sqrt{T}(\log T)^2 = o(H)$ and the following inequalities hold when $T\geqslant T_0(\varepsilon)$: Then there are a constant $c = c(\varepsilon)>0$ and a set $\mathcal{E}\subset (V,2V)$ of measure $\operatorname{mes}(\mathcal{E})>cV$ such that, for every $U\in \mathcal{E}$, we have Theorem 4. Under the hypotheses of Theorem 3 there are a constant $c = c(\varepsilon)>0$ and sets $\mathcal{E}_1, \mathcal{E}_2\subset (V,2V)$ such that $\operatorname{mes}(\mathcal{E}_j)>cV$ and for every $U_j\in \mathcal{E}_j$ we have Theorem 5. Suppose that $T\geqslant T_0$ and $D = D(T)\leqslant T^{1/6}$ is a monotone unbounded function of arbitrarily slow growth as $T\to +\infty$. Assume that $H = o(T)$, $\sqrt{T}(\log T)^2 = o(H)$, $U\leqslant T/(4D^2)$, and $\sqrt{T} = o(U)$. Then Theorem 6. Let $\delta$ be an arbitrary constant with $0<\delta<1/6$. Suppose that $T\geqslant T_0(\delta)$ and $\sqrt{T}(\log T)^3\leqslant H\leqslant T^{1-2\delta}$. Then there is a $U$ such that $\sqrt{T}\log{T}\leqslant U\leqslant T^{1/2+\delta}$ and where the constant implied by the $O$-symbol is absolute. This implies the existence of values of $U$ such that the quantity $\mathcal{Q}_{P}\,{=}\,\mathcal{Q}_{P}(T;\kern -0.3mm H,\kern -0.3mm U)$ is ‘anomalously’ small. More precisely, the following assertion holds. Corollary 1. The quantities $T$, $H$ and $U$ in Theorem 6 satisfy the bound where the constant implied by the $\ll$-symbol is absolute. The asymptotic formula (8) yields the existence of an unbounded sequence $t_{k}$, $k = 1,2,\dots$ such that $|P(t_{k})|\gg t_{k}^{1/4}$. In other words, the bound $P(t) = o(t^{1/4})$ as $t\to +\infty$ does not hold.4 However, the second corollary of Theorem 6 says that there are (sufficiently large) values of $U$ such that the difference between $P(t)$ and $P(t+U)$ does not exceed $o(t^{1/4})$ for ‘almost all’ $t$ in the interval $T\leqslant t\leqslant T+H$. Corollary 2. Let $T$, $H$ and $U$ be as in Theorem 6. Then the following equality holds for all $t$ in the interval $T\leqslant t\leqslant T+H$ outside a set of measure $\ll H(\log\log{T})^{-1/2}$: Notation. $\theta, \theta_1, \theta_2,\dots, \theta',\theta_1',\dots$ are complex numbers (different in different formulae) whose absolute values do not exceed 1, $c_1, c_2, \dots$ are absolute constants (different in different formulae) and $h(n)$ is the multiplicative function defined by the formula where $\chi_4$ is the non-principal Dirichlet character modulo $4$.§ 2. Lemmas on trigonometric integralsLet $T$, $H$, $U$, $N$ be such that
$$
\begin{equation*}
T\geqslant T_0,\qquad H,U = o(T),\qquad \sqrt{T}(\log T)^2=o(H),\qquad N = \frac{T}{4U}.
\end{equation*}
\notag
$$
Assume also that $\lambda,\mu>0$. We put
$$
\begin{equation*}
\begin{gathered} \, \kappa_1(\lambda)=\int_{T}^{T+H}\sqrt{t}\,e^{2\pi i\lambda\sqrt{t}}\,dt,\qquad \kappa_2(\lambda)=\int_{T}^{T+H}\sqrt{t}\,e^{2\pi i\lambda\sqrt{t+U}}\,dt, \\ \omega_1(\lambda,\mu)=\int_{T}^{T+H}\sqrt{t}\,e^{2\pi i(\lambda\sqrt{t+U}+\mu\sqrt{t})}\,dt, \\ \omega_2(\lambda,\mu)=\int_{T}^{T+H}\sqrt{t}\,e^{2\pi i(\lambda\sqrt{t+U}-\mu\sqrt{t})}\,dt. \end{gathered}
\end{equation*}
\notag
$$
Integrals of this type arise in the process of calculating Jutila’s integrals $\mathcal{Q}_{\Delta}$, $\mathcal{Q}_{P}$. Estimates for them were found in [2] for some values of the parameters $\lambda$ and $\mu$. Since [2] is not easily accessible and for the reader’s convenience, we shall give proofs of the estimates for $\kappa_j(\lambda)$ and $\omega_j(\lambda,\mu)$ which are different from the proofs in [2] and use only the second mean-value theorem (see, for example, [6], § 306). Proof. Under the change of variable $v = \lambda\sqrt{t}$, the integral $\kappa_1(\lambda)$ takes the form
$$
\begin{equation*}
\frac{2}{\lambda^3}\int_{v_1}^{v_2}g_1(v)e^{2\pi iv}\,dv,\qquad g_1(v)=v^2,\quad v_1=\lambda\sqrt{T},\quad v_2=\lambda\sqrt{T+H}.
\end{equation*}
\notag
$$
Using the second mean-value theorem for the real and imaginary parts of this integral, we obtain the following equalities for some $v_3$, $v_4$ such that $v_1\leqslant v_2$, $v_3\leqslant v_4$:
$$
\begin{equation*}
\begin{aligned} \, \kappa_1(\lambda) &=\frac{2}{\lambda^3}g_1(v_2)\biggl(\int_{v_1}^{v_3}\cos(2\pi v)\, dv+i\int_{v_1}^{v_4}\sin(2\pi v)\, dv\biggr) \\ &=\frac{g_1(v_2)}{\pi\lambda^3}\bigl(\sin(2\pi v_3)-\sin(2\pi v_1)+i(\cos(2\pi v_1)-\cos(2\pi v_4))\bigr) \\ &=\frac{g_1(v_2)}{\pi\lambda^3}\bigl(ie^{2\pi iv_1}+\sin(2\pi v_3)-i\cos(2\pi v_4)\bigr). \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
|\kappa_1(\lambda)|\leqslant \frac{g_1(v_2)}{\pi\lambda^3}(1+\sqrt{2}) =\frac{1+\sqrt{2}}{\pi}\,\frac{T+H}{\lambda}<\frac{4T}{5\lambda}.
\end{equation*}
\notag
$$
The same change of variable enables us to write the integral
$$
\begin{equation*}
\kappa_2(\lambda)=\int_{T_1}^{T_1+H}\sqrt{t-U}\,e^{2\pi i\lambda\sqrt{t}}\,dt,\qquad T_1=T+U,
\end{equation*}
\notag
$$
in the form Since $g_2(v)$ is increasing, the second mean-value theorem yields that Proof. We make the change of variable $v = v(t) = \lambda\sqrt{t+U}+\mu\sqrt{t}$. Then
$$
\begin{equation*}
\frac{dv}{dt}=\frac{\lambda\sqrt{t}+\mu\sqrt{t+U}}{2\sqrt{t}\sqrt{t+U}},\qquad \frac{dt}{dv}=\frac{2\sqrt{t}\sqrt{t+U}}{\lambda\sqrt{t}+\mu\sqrt{t+U}},
\end{equation*}
\notag
$$
so that $\omega_1(\lambda,\mu)$ takes the form
$$
\begin{equation*}
\begin{gathered} \, \int_{v_1}^{v_2}f(v)e^{2\pi iv}\,dv, \\ f(v)=\sqrt{t}\,\frac{dt}{dv}=\frac{2t\sqrt{t+U}}{\lambda\sqrt{t}+\mu\sqrt{t+U}},\qquad v_1 = v(T),\qquad v_2 = v(T+H). \end{gathered}
\end{equation*}
\notag
$$
It can easily be verified that
$$
\begin{equation*}
\begin{aligned} \, \frac{df}{dt} &=\frac{2\lambda\sqrt{t}(t+0.5U)+2\mu(t+U)^{3/2}}{\sqrt{t+U}(\lambda\sqrt{t}+\mu\sqrt{t+U})^2}, \\ \frac{df}{dv} &=\frac{df}{dt}\frac{dt}{dv}=4\sqrt{t}\, \frac{\lambda\sqrt{t}(t+0.5U)+\mu(t+U)^{3/2}}{(\lambda\sqrt{t}+\mu\sqrt{t+U})^3}. \end{aligned}
\end{equation*}
\notag
$$
Hence $f(v)$ is an increasing function of $v$. Using the second mean-value theorem, we find that
$$
\begin{equation*}
\begin{aligned} \, |\omega_1(\lambda,\mu)| &\leqslant \frac{f(v_2)}{2\pi}\,(1+\sqrt{2})=\frac{1+\sqrt{2}}{2\pi}\, \frac{2(T+H)\sqrt{T+H+U}}{\lambda\sqrt{T+H}+\mu\sqrt{T+H+U}} \\ &<\frac{1+\sqrt{2}}{\pi}\,\frac{\sqrt{T+H}\sqrt{T+H+U}}{\lambda+\mu}<\frac{4T}{5(\lambda+\mu)}, \end{aligned}
\end{equation*}
\notag
$$
as required. $\Box$ Lemma 3. Let $m,n$ be integers with $1\leqslant m<n$ and $m\leqslant N$, and let $(\lambda,\mu)$ be either of the pairs $(\sqrt{m},\sqrt{n})$, $(\sqrt{n},\sqrt{m})$. Then Proof. We substitute $v = v(t) = \lambda\sqrt{t+U}-\mu\sqrt{t}$ and put
$$
\begin{equation*}
u_1 = v(T), \qquad u_2 = v(T+H), \qquad v_1 = \min{(u_1,u_2)}, \qquad v_2 = \max{(u_1,u_2)}.
\end{equation*}
\notag
$$
Under this change of variable, the integral takes the form
$$
\begin{equation*}
\omega_2(\lambda,\mu)=\int_{u_1}^{u_2}g(v)e^{2\pi iv}dv,\qquad g(v)=\sqrt{t}\,\frac{dt}{dv}=\frac{2\sqrt{t}\sqrt{t+U}}{\lambda\sqrt{t}-\mu\sqrt{t+U}},\quad t = t(v).
\end{equation*}
\notag
$$
An expression for the derivative $dg/dt$ can be obtained from that found in Lemma 2 by taking the opposite sign of $\mu$. Hence,
$$
\begin{equation*}
\frac{dg}{dv}=4\sqrt{t}\,\frac{P}{Q^3},\quad\text{where}\quad P = \lambda\sqrt{t}(t+0.5U)-\mu(t+U)^{3/2},\quad Q = \lambda\sqrt{t}-\mu\sqrt{t+U}.
\end{equation*}
\notag
$$
We claim that $g(v)$ is an increasing function of $v$. Indeed, it suffices to show that $P$ and $Q$ are of the same sign. Suppose that $\lambda = \sqrt{m}$, $\mu = \sqrt{n}$. Then, for every $t$, $T\leqslant t\leqslant T+H$, we have
$$
\begin{equation}
-Q=\mu\sqrt{t+U}-\lambda\sqrt{t}\geqslant \sqrt{t}(\mu-\lambda)\geqslant \sqrt{T}(\sqrt{n}-\sqrt{m})>0,\qquad Q<0.
\end{equation}
\tag{13}
$$
In a similar vein,
$$
\begin{equation*}
\begin{aligned} \, -P &=\mu(t+U)^{3/2}-\lambda\sqrt{t}(t+0.5U)>\mu(t+U)^{3/2}-\lambda\sqrt{t}(t+U) \\ &=-(t+U)Q>0,\qquad P<0. \end{aligned}
\end{equation*}
\notag
$$
We now suppose that $\lambda = \sqrt{n}$, $\mu = \sqrt{m}$. Then
$$
\begin{equation}
\begin{aligned} \, Q &=\sqrt{nt}-\sqrt{m(t+U)}=\frac{nt-m(t+U)}{\sqrt{nt}+\sqrt{m(t+U)}} =\frac{(n-m)t-mU}{\sqrt{nt}+\sqrt{m(t+U)}} \nonumber \\ &\geqslant \frac{T(n-m)-NU}{\sqrt{nt}+\sqrt{m(t+U)}}\geqslant \frac{T(n-m-1/4)}{(\sqrt{n}+\sqrt{m})\sqrt{T+H+U}} \nonumber \\ &\geqslant \frac{T}{4\sqrt{T+H+U}}\,\frac{n-m}{\sqrt{n}+\sqrt{m}} \geqslant \frac{\sqrt{T}}{5}(\sqrt{n}-\sqrt{m}). \end{aligned}
\end{equation}
\tag{14}
$$
Furthermore,
$$
\begin{equation*}
P=\sqrt{nt}(t+0.5U)-\sqrt{m}(t+U)^{3/2} =\frac{nt(t+0.5U)^2-m(t+U)^3}{\sqrt{nt}(t+0.5U)+\sqrt{m}(t+U)^{3/2}}.
\end{equation*}
\notag
$$
Putting $\delta = U/t$, we can represent the numerator as the product $t^3R$, where $R = n(1+0.5\delta)^2-m(1+\delta)^3$. Since
$$
\begin{equation*}
\delta \leqslant \frac{U}{T}=\frac{1}{4N}\leqslant \frac{1}{4m}\leqslant \frac{1}{4},
\end{equation*}
\notag
$$
we have
$$
\begin{equation*}
\begin{aligned} \, R &=n-m+(\delta n - 3\delta m)+\biggl(\frac{\delta^2}{4}n-\delta^3m\biggr)-3\delta^2m\geqslant n-m-2\delta m-3\delta^2m \\ &\geqslant n-m-3\delta m\geqslant n-m-\frac{3}{4}\geqslant \frac{1}{4}(n-m)>0. \end{aligned}
\end{equation*}
\notag
$$
Therefore, $P>0$. Thus $g(v)$ is an increasing function of $v$. It follows from the inequalities (13), (14) that the derivative
$$
\begin{equation*}
\frac{dv}{dt}=\frac{\lambda\sqrt{t}-\mu\sqrt{t+U}}{2\sqrt{t}\sqrt{+U}}
\end{equation*}
\notag
$$
is negative when $\lambda = \sqrt{m}$, $\mu = \sqrt{n}$ and positive when $\lambda = \sqrt{n}$, $\mu = \sqrt{m}$. Therefore, putting $\varepsilon = -1$ in the first case and $\varepsilon = 1$ in the second, we find that
$$
\begin{equation*}
\omega_2(\lambda,\mu)=\varepsilon\int_{v_1}^{v_2}g(v)e^{2\pi iv}\, dv.
\end{equation*}
\notag
$$
Using the second mean-value theorem for the real and imaginary parts of this integral and noticing that we obtain as required. $\Box$ § 3. An explicit formula for $\mathcal{K}_{P}$ and some consequencesIn this section we derive an analogue of (4) for Jutila’s integral. Our derivation is based on the so-called truncated Hardy–Voronoï formula (Lemma 4) and estimates for certain sums containing the function $r(n)$ (Lemma 5). We use Chamizo’s deep result on the mean-value of $r(n)r(n+k)$ (Lemma 6) to improve the power of the logarithm in the estimate of a sum in Lemma 7 and, as a corollary, in the estimates of the error terms in Lemmas 8, 10–12. This in turn influences the lower bound for the parameter $H$ in the corollary of Lemma 8 and in subsequent assertions.5 We also prove an asymptotic formula for the second moment of $P(t)$ on the interval $T\leqslant t\leqslant T+H$ (Lemma 8), which is needed in the derivation of the explicit formula for the correlation function $\mathcal{K}_{P}$ (Lemmas 12, 13). Lemma 4. For every $N\geqslant 1$ we have the following formula as $t\to +\infty$: where Proof. This formula is proved by the argument used for the truncated Voronoï formula for $\Delta(x)$ in [7], § 3.12. The error term in the latter formula is of order $N^{\varepsilon} + t^{\varepsilon}\sqrt{t/N}$. But the argument can be improved as desired in view of the inequality which holds for all $c>\log{2}$. $\Box$ Lemma 5. As $X\to +\infty$, we have Proof. The first relation was obtained (in a sharpened form) by Sierpiński in 1908 [8] (see also [9]). The first two inequalities can be obtained from it using Abel transformation. To prove the two last bounds, we consider the following Dirichlet series for $\operatorname{Re} s >1$:
$$
\begin{equation*}
F(s)=\sum_{n=1}^{+\infty}\frac{h^2(n^2)}{n^{s}}=\prod_{p}F_{p}(s),\qquad F_{p}(s)=1+\frac{h^2(p^{2s})}{p^{s}}+\frac{h^2(p^{2s})}{p^{2s}} +\frac{h^2(p^{6s})}{p^{3s}}+\cdots\,.
\end{equation*}
\notag
$$
An easy verification shows that
$$
\begin{equation*}
\begin{aligned} \, F_{p}(s) &=\biggl(1-\frac{1}{p^{s}}\biggr)^{-1}\quad\text{when}\quad p\not\equiv 1\pmod{4}, \\ F_{p}(s) &=1+\frac{3^2}{p^{s}}+\frac{5^2}{p^{2s}}+\frac{7^2}{p^{3s}}+\cdots= \frac{1+6p^{-s}+p^{-2s}}{(1-p^{-s})^3} =\biggl(1-\frac{1}{p^{s}}\biggr)^{-9}\biggl(1-\frac{20}{p^{2s}} \\ &\qquad+\frac{64}{p^{3s}}-\frac{90}{p^{4s}}+\frac{64}{p^{5s}}-\frac{20}{p^{6s}} +\frac{1}{p^{8s}}\biggr) \quad\text{when}\quad p\equiv 1\pmod{4}. \end{aligned}
\end{equation*}
\notag
$$
We can conclude that $F(s) = \zeta^5(s)\Phi(s)$, where $\Phi(s)$ is analytic for $\operatorname{Re} s > 1/2$. A standard application of the method of complex integration now yields the asymptotic equality where $P_4(u)$ is a quartic polynomial, $0<c<1$. The required bounds can be obtained from this formula by Abel transformation. $\Box$ Lemma 6. Let $k$, $N$ be integers with $1\leqslant k \ll N$. Suppose that $2^r\parallel k$. Then where $\sigma(d)$ is the sum of the divisors of $d$ and the constant implied by the $\ll$-symbol is absolute. This is Corollary 5.3 in [10]. Lemma 7. Suppose that $\nu\geqslant 3$. Then we have where the constant implied by the $\ll$-symbol is absolute. Proof. Denoting the sum by $S(\nu)$, we have
$$
\begin{equation*}
S(\nu)\ll \sum_{1\leqslant m<n\leqslant \nu}\frac{r(m)}{m^{3/4}}\,\frac{r(n)}{n^{1/4}}\,\frac{1}{n-m}.
\end{equation*}
\notag
$$
Let $S_1(\nu)$ be the contribution of all $m\leqslant n/2$ to the right-hand side. Clearly,
$$
\begin{equation*}
S_1(\nu)\ll \sum_{n\leqslant \nu}\frac{r(n)}{n^{5/4}}\sum_{m\leqslant n/2}\frac{r(m)}{m^{3/4}}\ll \sum_{m\leqslant \nu}\frac{r(n)}{n}\ll \log{\nu}.
\end{equation*}
\notag
$$
Furthermore, let $S_2(\nu)$ be the contribution of the summands with $n/2<m\leqslant n-1$. Putting $n = m+k$, we have
$$
\begin{equation*}
\begin{aligned} \, S_2(\nu) &\ll \sum_{1\leqslant m\leqslant \nu}\,\sum_{m+1\leqslant n\leqslant 2m}\frac{r(m)r(n)}{m(n-m)}\ll \sum_{1\leqslant m\leqslant \nu}\sum_{k=1}^{m}\frac{1}{km}\,r(m)r(m+k) \\ &\ll \sum_{1\leqslant k\leqslant \nu}\frac{1}{k}\sum_{k\leqslant m\leqslant \nu}\frac{1}{m}\,r(m)r(m+k). \end{aligned}
\end{equation*}
\notag
$$
We split the inner sum into intervals of the form $M<m\leqslant M_1$, where $M_1\leqslant 2M$, $k\leqslant M\leqslant \nu/2$. Then, in view of Lemma 6,
$$
\begin{equation*}
\begin{aligned} \, S_2(\nu) &\ll \sum_{k\leqslant \nu}\frac{1}{k}\mathop{{\sum}'}_{M}\frac{1}{M}\sum_{M<m\leqslant M_1}r(m)r(m+k)\ll \sum_{k\leqslant \nu}\frac{1}{k}\mathop{{\sum}'}_{M}\frac{1}{M}\,\frac{2^r}{k} \, \sigma\biggl(\frac{k}{2^r}\biggr)M \\ &\ll (\log{\nu})\sum_{k\leqslant \nu}\frac{2^r}{k^2}\, \sigma\biggl(\frac{k}{2^r}\biggr), \end{aligned}
\end{equation*}
\notag
$$
where $2^r\parallel k$. Putting $k = 2^rt$, where $t$ is odd, we find that Proof. This integral can be rewritten in the form
$$
\begin{equation*}
\begin{aligned} \, &\int_{T}^{T+H}\sqrt{t}\sum_{m,n\leqslant T}\frac{r(m)r(n)}{(mn)^{3/4}}e^{2\pi i(\sqrt{n}-\sqrt{m})\sqrt{t}}\,dt \\ &\qquad=\biggl(\sum_{n\leqslant T}\frac{r^2(n)}{n^{3/2}}\biggr) \int_{T}^{T+H}\sqrt{t}\,dt+\sum_{\substack{m,n\leqslant T \\ m\ne n}}\frac{r(m)r(n)}{(mn)^{3/4}}\kappa_1(\sqrt{n}-\sqrt{m}). \end{aligned}
\end{equation*}
\notag
$$
By Lemmas 1 and 7, the order of the last sum does not exceed
$$
\begin{equation*}
T\sum_{m<n\leqslant T}\frac{r(m)r(n)}{(mn)^{3/4}}\,\frac{1}{\sqrt{n}-\sqrt{m}}\ll T(\log T)^2.
\end{equation*}
\notag
$$
By Lemma 5, and the lemma follows. $\Box$ Corollary 3. Suppose that $T\to +\infty$, $H = o(T)$, $\sqrt{T}(\log T)^2 = o(H)$. Then the following asymptotic formula holds: Lemma 9. Let $T\to +\infty$, $\varpi(T) = (\log{T})^{3/2}\log\log{T}$, and suppose that $H = o(T)$, $\sqrt{T}\,\varpi(T) = o(H)$. Then Proof. The desired relation is a direct corollary of the formula
$$
\begin{equation*}
\int_0^{T}P^2(t)\, dt=\frac{A}{3\pi^2}T^{3/2}+O\bigl(T\varpi(T)\bigr)
\end{equation*}
\notag
$$
contained in [4]. $\Box$ Lemma 10. Let $T\to +\infty$, $H,U = o(T)$, $\sqrt{T}(\log T)^2 = o(H)$. Then we have where Proof. Put $\psi(t) = (t+U)^{1/4}-t^{1/4}$. Then $\psi(t)\ll UT^{-3/4}$ when $T\leqslant t\leqslant T+H$. Hence, by Lemma 4, we have
$$
\begin{equation*}
\begin{aligned} \, &P(t+U)-P(t) =-\frac{1}{\pi}(t^{1/4}+\psi(t))\sum_{n\leqslant T}\frac{r(n)}{n^{3/4}} \cos\biggl(2\pi\sqrt{n}\sqrt{t+U}+\frac{\pi}{4}\biggr) \\ &\qquad\qquad+\frac{t^{1/4}}{\pi}\sum_{n\leqslant T}\frac{r(n)}{n^{3/4}}\cos\biggl(2\pi\sqrt{n}\sqrt{t}+\frac{\pi}{4}\biggr)+ O\bigl(\varphi(T)\bigr) \\ &\qquad=-\frac{t^{1/4}}{\pi}\sum_{n\leqslant T}\frac{r(n)}{n^{3/4}} \biggl(\cos\biggl(2\pi\sqrt{n}\sqrt{t+U}+\frac{\pi}{4}\biggr) -\cos\biggl(2\pi\sqrt{n}\sqrt{t}+\frac{\pi}{4}\biggr)\biggr) \\ &\qquad\qquad-\frac{\psi(t)}{\pi}\sum_{n\leqslant T} \frac{r(n)}{n^{3/4}} \cos\biggl(2\pi\sqrt{n}\sqrt{t+U}+\frac{\pi}{4}\biggr)+O\bigl(\varphi(T)\bigr) \\ &\qquad=-(\mathcal{A}(t)+\mathcal{B}(t)+\mathcal{C}(t)), \end{aligned}
\end{equation*}
\notag
$$
where the meaning of the notation is clear. Accordingly,
$$
\begin{equation*}
\begin{aligned} \, &\mathcal{Q}_{P}(T;H,U) \\ &\qquad=\int_{T}^{T+H}\bigl(\mathcal{A}^2(t)+\mathcal{B}^2(t)+\mathcal{C}^2(t)+ 2\mathcal{A}(t)\mathcal{B}(t)+2\mathcal{A}(t)\mathcal{C}(t) +2\mathcal{B}(t)\mathcal{C}(t)\bigr)\,dt \\ &\qquad=\mathcal{Q}^{(1)}+\dots+\mathcal{Q}^{(6)}. \end{aligned}
\end{equation*}
\notag
$$
Clearly, $|\mathcal{Q}^{(4)}|\leqslant 2\sqrt{\mathcal{Q}^{(1)}\mathcal{Q}^{(2)}}$, $|\mathcal{Q}^{(5)}|\leqslant 2\sqrt{\mathcal{Q}^{(1)}\mathcal{Q}^{(3)}}$, $|\mathcal{Q}^{(6)}|\leqslant 2\sqrt{\mathcal{Q}^{(2)}\mathcal{Q}^{(3)}}$. Furthermore, $\mathcal{Q}^{(3)}\ll H\varphi^2(T)$. By Corollary 3,
$$
\begin{equation*}
\begin{aligned} \, \mathcal{Q}^{(2)} &\leqslant \frac{1}{\pi}\int_{T}^{T+H}\psi^2(t)\biggl|\sum_{n\leqslant T}\frac{r(n)}{n^{3/4}}\cos\biggl(2\pi\sqrt{n}\sqrt{t+U}+\frac{\pi}{4}\biggr)\biggr|^2\, dt \\ &\ll \frac{U^2}{T^{3/2}}\int_{T}^{T+H}\biggl|\sum_{n\leqslant T} \frac{r(n)}{n^{3/4}} \cos\biggl(2\pi\sqrt{n}\sqrt{t+U}+\frac{\pi}{4}\biggr)\biggr|^2\, dt \\ &\ll \frac{U^2}{T^2}\int_{T+U}^{T+H+U}\sqrt{t}\,\biggl|\sum_{n\leqslant T} \frac{r(n)}{n^{3/4}}\cos\biggl(2\pi\sqrt{n}\sqrt{t}+\frac{\pi}{4}\biggr)\biggr|^2\, dt \\ &\ll \frac{U^2}{T^2}\int_{T+U}^{T+U+H}\sqrt{t}\,\biggl| \sum_{n\leqslant T} \frac{r(n)}{n^{3/4}}e^{2\pi i\sqrt{n}\sqrt{t}}\biggr|^2\, dt\ll \frac{U^2}{T^2}H\sqrt{T}. \end{aligned}
\end{equation*}
\notag
$$
Moreover, a rough estimate for $\mathcal{Q}^{(1)}$ yields that
$$
\begin{equation*}
\begin{aligned} \, \mathcal{Q}^{(1)} &\leqslant \frac{1}{\pi^2}\int_{T}^{T+H}\!\! \sqrt{t}\,\biggl|\sum_{n\leqslant T} \frac{r(n)}{n^{3/4}}\bigl(e^{2\pi i\sqrt{n}\sqrt{t+U}}-e^{2\pi i\sqrt{n}\sqrt{t}}\bigr)\biggr|^2\, dt \\ &\leqslant\frac{2}{\pi^2}\int_{T}^{T+H}\!\! \sqrt{t}\,\biggl(\biggl|\sum_{n\leqslant T}\frac{r(n)}{n^{3/4}}e^{2\pi i\sqrt{n}\sqrt{t+U}}\biggr|^2+\biggl|\sum_{n\leqslant T}\frac{r(n)}{n^{3/4}}e^{2\pi i\sqrt{n}\sqrt{t}}\biggr|^2\biggr)\, dt\ll H\sqrt{T}. \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\begin{gathered} \, \mathcal{Q}^{(4)}\ll H\sqrt{T}\,\frac{U}{T},\qquad \mathcal{Q}^{(5)}\ll H\sqrt{T}\,\frac{\varphi(T)}{\sqrt[4]{T}}, \\ \mathcal{Q}^{(6)}\ll H\sqrt{T}\,\frac{U}{T}\frac{\varphi(T)}{\sqrt[4]{T}}\ll H\sqrt{T}\,\frac{\varphi(T)}{\sqrt[4]{T}}, \end{gathered}
\end{equation*}
\notag
$$
whence we have
$$
\begin{equation*}
\mathcal{Q}_{P}=\mathcal{Q}^{(1)}+O\biggl(H\sqrt{T}\biggl(\frac{U}{T} +\frac{\varphi(T)}{\sqrt[4]{T}}\biggr)\biggr).
\end{equation*}
\notag
$$
We put $N = T(4U)^{-1}$ and introduce the notation
$$
\begin{equation*}
\begin{aligned} \, W(t) &=\frac{e^{\pi i/4}}{\pi}\sum_{n\leqslant T}\frac{r(n)}{n^{3/4}}\bigl(e^{2\pi i\sqrt{n}\sqrt{t+U}}-e^{2\pi i\sqrt{n}\sqrt{t}}\bigr) \\ &=\frac{e^{\pi i/4}}{\pi}\biggl(\sum_{n\leqslant N}+\sum_{N<n\leqslant T}\biggr)\frac{r(n)}{n^{3/4}}\bigl(e^{2\pi i\sqrt{n}\sqrt{t+U}}-e^{2\pi i\sqrt{n}\sqrt{t}}\bigr)=W_1(t)+W_2(t). \end{aligned}
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\begin{aligned} \, \operatorname{Re}^2W(t) &=\biggl\{\frac{1}{2}(W_1+W_2+\overline{W}_1+\overline{W}_2)\biggr\}^2 =\frac{1}{4}(W_1+\overline{W}_1+W_2+\overline{W}_2)^2 \\ &=\frac{1}{2}|W_1|^2+\frac{1}{2}\operatorname{Re}(W_1^2)+\operatorname{Re}{(W_1W_2)} +\operatorname{Re}{(W_1\overline{W}_2)}+\operatorname{Re}^2(W_2). \end{aligned}
\end{equation*}
\notag
$$
It follows that $\mathcal{Q}^{(1)} = q_P/2+q_1/2+q_2+q_3+q_4$, where the meaning of the notation is clear.
First of all, we have
$$
\begin{equation*}
\begin{aligned} \, q_1 &=\operatorname{Re} \frac{i}{\pi^2}\int_{T}^{T+H}\sqrt{t}\,\sum_{m\leqslant N<n\leqslant T}\frac{r(m)r(n)}{(mn)^{3/4}} \\ &\qquad\qquad\times\bigl(e^{2\pi i\sqrt{m}\sqrt{t+U}}-e^{2\pi i\sqrt{m}\sqrt{t}}\bigr) \bigl(e^{2\pi i\sqrt{n}\sqrt{t+U}}-e^{2\pi i\sqrt{n}\sqrt{t}}\bigr)\, dt \\ &=-\frac{1}{\pi^2}\operatorname{Im}\sum_{m,n\leqslant N}\frac{r(m)r(n)}{(mn)^{3/4}} \\ &\qquad\qquad\times\bigl(\kappa_1(\sqrt{m}+\sqrt{n})+\kappa_2(\sqrt{m}+\sqrt{n}) -\omega_1(\sqrt{m},\sqrt{n})-\omega_1(\sqrt{n},\sqrt{m})\bigr). \end{aligned}
\end{equation*}
\notag
$$
Using Lemma 1, we find that
$$
\begin{equation*}
q_1\ll T\sum_{m,n\leqslant T}\frac{r(m)r(n)}{(mn)^{3/4}}\,\frac{1}{\sqrt{m}+\sqrt{n}}\ll T\log{T}.
\end{equation*}
\notag
$$
An estimate for the integral $q_2$ can be obtained in a similar way:
$$
\begin{equation*}
\begin{aligned} \, q_2 &=\operatorname{Re} \frac{i}{\pi^2}\int_{T}^{T+H}\sqrt{t}\,\sum_{m\leqslant N<n\leqslant T}\frac{r(m)r(n)}{(mn)^{3/4}} \\ &\qquad\qquad\times\bigl(e^{2\pi i\sqrt{m}\sqrt{t+U}}-e^{2\pi i\sqrt{m}\sqrt{t}}\bigr) \bigl(e^{2\pi i\sqrt{n}\sqrt{t+U}}-e^{2\pi i\sqrt{n}\sqrt{t}}\bigr)\, dt \\ &=-\frac{1}{\pi^2}\operatorname{Im}\sum_{m\leqslant N<n\leqslant T} \frac{r(m)r(n)}{(mn)^{3/4}} \\ &\qquad\qquad\times\bigl(\kappa_1(\sqrt{m}+\sqrt{n})+\kappa_2(\sqrt{m}+\sqrt{n}) -\omega_1(\sqrt{m},\sqrt{n})-\omega_1(\sqrt{n},\sqrt{m})\bigr) \\ &\ll T\log{T}. \end{aligned}
\end{equation*}
\notag
$$
Furthermore, using the bounds in Lemmas 3 and 7, we conclude that
$$
\begin{equation*}
\begin{aligned} \, q_3 &=\frac{1}{\pi^2}\operatorname{Re} \int_{T}^{T+H}\sqrt{t}\,\sum_{m\leqslant N<n\leqslant T}\frac{r(m)r(n)}{(mn)^{3/4}} \\ &\qquad\qquad\times\bigl(e^{2\pi i\sqrt{m}\sqrt{t+U}}-e^{2\pi i\sqrt{m}\sqrt{t}}\bigr) \bigl(e^{-2\pi i\sqrt{n}\sqrt{t+U}}-e^{-2\pi i\sqrt{n}\sqrt{t}}\bigr)\, dt \\ &=\frac{1}{\pi^2}\operatorname{Re} \sum_{m\leqslant N<n\leqslant T}\frac{r(m)r(n)}{(mn)^{3/4}} \\ &\qquad\qquad\times\bigl(\overline{\kappa}_1(\sqrt{n}-\sqrt{m})+\overline{\kappa}_2(\sqrt{n}-\sqrt{m})- \omega_2(\sqrt{m},\sqrt{n})-\overline{\omega}_2(\sqrt{n},\sqrt{m})\bigr) \\ &\ll T\sum_{m\leqslant N<n\leqslant T}\frac{r(m)r(n)}{(mn)^{3/4}}\,\frac{1}{\sqrt{n}-\sqrt{m}}\ll T(\log T)^2. \end{aligned}
\end{equation*}
\notag
$$
We now proceed to estimate $q_4$. We have
$$
\begin{equation*}
\begin{aligned} \, q_4 &\leqslant \frac{2}{\pi^2}\int_{T}^{T+H}\sqrt{t}\biggl(\biggl|\sum_{N<n\leqslant T}\frac{r(n)}{n^{3/4}}e^{2\pi i\sqrt{n}\sqrt{t+U}}\biggr|^2+\biggl|\sum_{N<n\leqslant T}\frac{r(n)}{n^{3/4}}e^{2\pi i\sqrt{n}\sqrt{t}}\biggr|^2\biggr)\, dt \\ &\leqslant \frac{4}{\pi^2}\int_{T_1}^{T_1+H}\sqrt{t}\biggl|\sum_{N<n\leqslant T}\frac{r(n)}{n^{3/4}}e^{2\pi i\sqrt{n}\sqrt{t}}\biggr|^2\, dt, \end{aligned}
\end{equation*}
\notag
$$
where $T_1$ is the one of $T$, $T+U$ for which the integral is maximal. Accordingly,
$$
\begin{equation*}
\begin{aligned} \, q_4 &\leqslant \frac{4}{\pi^2}\int_{T_1}^{T_1+H}\sqrt{t}\sum_{N<m,n\leqslant T}\frac{r(m)r(n)}{(mn)^{3/4}}e^{2\pi i(\sqrt{n}-\sqrt{m})\sqrt{t}}\, dt \\ &=\frac{4}{\pi^2}\biggl(\int_{T_1}^{T_1+H}\sqrt{t}\,dt\biggr)\sum_{N<n\leqslant T}\frac{r^2(n)}{n^{3/2}}+ \frac{4}{\pi^2}\sum_{\substack{N<m<n\leqslant T \\ m\ne n}}\frac{r(m)r(n)}{(mn)^{3/4}}\kappa_1(\sqrt{n}-\sqrt{m}) \\ &\ll H\sqrt{T}\,\frac{\log{N}}{\sqrt{N}}+T(\log T)^2\ll H\sqrt{U}\log{\frac{T}{U}}+T(\log T)^2. \end{aligned}
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
\sum_{k = 1}^4|q_{k}|\ll H\sqrt{U}\log{\frac{T}{U}}+T(\log T)^2\ll H\sqrt{T}\biggl(\frac{\sqrt{T}}{H}(\log T)^2+\sqrt{\frac{U}{T}}\log{\frac{T}{U}}\biggr),
\end{equation*}
\notag
$$
$$
\begin{equation*}
\begin{aligned} \, \mathcal{Q}^{(1)} &=\frac{1}{2\pi^2}\int_{T}^{T+H}\sqrt{t}\biggl|\sum_{n\leqslant N}\frac{r(n)}{n^{3/4}}\bigl(e^{2\pi i\sqrt{n}\sqrt{t+U}}-e^{2\pi i\sqrt{n}\sqrt{t}}\bigr)\biggr|^2\, dt \\ &\qquad+O\biggl(H\sqrt{T}\biggl(\frac{\sqrt{T}}{H}(\log T)^2 +\sqrt{\frac{U}{T}}\log{\frac{T}{U}}\biggr)\biggr), \\ \mathcal{Q}_{P} &=\frac{1}{2\pi^2}\int_{T}^{T+H}\sqrt{t}\biggl|\sum_{n\leqslant N}\frac{r(n)}{n^{3/4}}\bigl(e^{2\pi i\sqrt{n}\sqrt{t+U}}-e^{2\pi i\sqrt{n}\sqrt{t}}\bigr)\biggr|^2\, dt \\ &\qquad+O\biggl(H\sqrt{T}\biggl(\frac{\sqrt{T}}{H}(\log T)^2+\frac{U}{T} +\sqrt{\frac{U}{T}}\log{\frac{T}{U}}+\frac{\varphi(T)}{\sqrt[4]{T}}\biggr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
Noticing that we arrive at the desired assertion. $\Box$ Lemma 11. Under the hypotheses of Lemma 10, the integral $q_{P}$ satisfies the equality where Proof. We have
$$
\begin{equation*}
\begin{aligned} \, q_{P} &=\int_{T}^{T+H}\sqrt{t}\sum_{m,n\leqslant N}\frac{r(m)r(n)}{(mn)^{3/4}} \\ &\qquad\qquad\times \bigl(e^{2\pi i\sqrt{m}\sqrt{t+U}}-e^{2\pi i\sqrt{m}\sqrt{t}}\bigr) \bigl(e^{-2\pi i\sqrt{n}\sqrt{t+U}}-e^{-2\pi i\sqrt{n}\sqrt{t}}\bigr)\, dt \\ &=\sum_{n\leqslant N}\frac{r^2(n)}{n^{3/2}}\int_{T}^{T+H}\sqrt{t}\,\bigl|e^{2\pi i\sqrt{n}\sqrt{t+U}}-e^{2\pi i\sqrt{n}\sqrt{t}}\bigr|^2\, dt +\sum_{\substack{m,n\leqslant N \\ m\ne n}}\frac{r(m)r(n)}{(mn)^{3/4}} \\ &\qquad\qquad\times\bigl(\kappa_1(\sqrt{m}-\sqrt{n})+\kappa_2(\sqrt{m}-\sqrt{n}) -\omega_2(\sqrt{m},\sqrt{n})-\overline{\omega}_2(\sqrt{n},\sqrt{m})\bigr). \end{aligned}
\end{equation*}
\notag
$$
Estimating the contribution of the summands with $m\ne n$ using Lemmas 1–3, 7, we arrive at the desired relation. $\Box$ Corollary 4 (an analogue of Jutila’s formula for the integral $\mathcal{Q}_{P}$). Let $T{\to}\,{+}\infty$, $H,U = o(T)$, $\sqrt{T}(\log T)^2 = o(H)$. Then Note that the integral $I(n)$ (defined in (15)) is different from the integral
$$
\begin{equation*}
\mathfrak{I}(n) = \frac{1}{n^{3/2}}\int_{T}^{T+H}\sqrt{t}\,\bigl|e^{\pi iU\sqrt{n/t}}-1\bigr|^2\, dt
\end{equation*}
\notag
$$
in Jutila’s original formula (4). However, we can show that $\mathfrak{I}(n)$ approximates $I(n)$ rather closely when $U\leqslant T^{2/3-\varepsilon}$. Indeed, noticing that
$$
\begin{equation*}
\sqrt{t+U}-\sqrt{t}=\frac{U}{2\sqrt{t}} + O\biggl(\frac{U^2}{T^{3/2}}\biggr)
\end{equation*}
\notag
$$
when $T\leqslant t\leqslant T+H$ and using the inequality which holds for all complex numbers $u$, $v$, we find that
$$
\begin{equation*}
\begin{gathered} \, \bigl|e^{2\pi i\sqrt{n}(\sqrt{t+U}-\sqrt{t})}-1\bigr|^2 = \bigl|e^{\pi iU\sqrt{n/t}}-1\bigr|^2+O\biggl(\frac{U^2\sqrt{n}}{T^{3/2}}\biggr), \\ I(n)=\frac{1}{n^{3/2}}\int_{T}^{T+H}\sqrt{t}\biggl(\bigl|e^{\pi iU\sqrt{n/t}}-1\bigr|^2+O\biggl(\frac{U^2\sqrt{n}}{T^{3/2}}\biggr)\biggr)\, dt=\mathfrak{I}(n)+O\biggl(\frac{HU^2}{nT}\biggr). \end{gathered}
\end{equation*}
\notag
$$
If $U\gg \sqrt{T}$, then $I(n)\asymp {H\sqrt{T}}/{n^{3/2}}$ (this follows from the proof of Lemma 12). Therefore
$$
\begin{equation*}
\frac{HU^2}{nT} \asymp I(n)\frac{U^2\sqrt{n}}{T^{3/2}}\ll I(n)\frac{U^2}{T^{3/2}}\sqrt{\frac{T}{U}}\ll I(n)\frac{U^{3/2}}{T}\ll I(n)T^{-3\varepsilon/2},
\end{equation*}
\notag
$$
whence we have
$$
\begin{equation*}
\mathfrak{I}(n) =I(n)\bigl(1+O(T^{-3\varepsilon/2})\bigr),\qquad I(n)=\mathfrak{I}(n)\bigl(1+O(T^{-3\varepsilon/2})\bigr).
\end{equation*}
\notag
$$
But if $U\leqslant 0.5\sqrt{T}$, then $\mathfrak{I}(n)\asymp (HU/n^{3/2})\log({\sqrt{T}}/{U})$ by the results in [2] and, therefore,
$$
\begin{equation*}
\frac{HU^2}{nT} \ll \mathfrak{I}(n)\frac{U\sqrt{n}}{T}\ll \mathfrak{I}(n)\frac{U}{T}\sqrt{\frac{T}{U}}\ll \mathfrak{I}(n)\sqrt{\frac{U}{T}}\ll \mathfrak{I}(n)T^{-0.25},
\end{equation*}
\notag
$$
whence For these values of $U$, the formula in the corollary yields that
$$
\begin{equation*}
\mathcal{Q}_{P}(T,H;U)=\frac{1}{2\pi^2}\sum_{n\leqslant N}r^2(n)\mathfrak{I}(n)+O(R),
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\begin{aligned} \, R &\ll H\sqrt{T}\biggl(\frac{\sqrt{T}}{H}(\log T)^2+\sqrt{\frac{U}{T}}\log{\frac{T}{U}} +\frac{\varphi(T)}{\sqrt[4]{T}}+\frac{U^2}{T^{3/2}}(\log T)^2\biggr) \\ &\ll H\sqrt{T}\biggl(\frac{\sqrt{T}}{H}(\log T)^2+\sqrt{\frac{U}{T}}\log{\frac{T}{U}} +\frac{\varphi(T)}{\sqrt[4]{T}}\biggr)\ll T^{\varepsilon}\bigl(T+H\sqrt{U}+H\sqrt[4]{T}\bigr). \end{aligned}
\end{equation*}
\notag
$$
If $U\,{\ll}\,\sqrt{T}$, then using the inequality $\mathcal{Q}^{(1)}\ll HU\log{(\sqrt{T}/U)}$ instead of $\mathcal{Q}^{(1)}\ll H\sqrt{T}$ in the proof of Lemma 10 enables us to get rid of the summand $H\sqrt[4]{T}$ in the estimate for $R$ and arrive exactly at Jutila’s formula (4). In what follows we always assume that $T$ is large, $T\to +\infty$, $U = o(T)$, $N = T(4U)^{-1}$. Let $F(n)$ be an arbitrary sequence of numbers ($1\,{\leqslant}\, n\,{\leqslant}\,N$). We put To work with the sum $\Sigma_{I}$, where $I = I(n)$ are the numbers defined in Lemma 11, we need the following lemma. Lemma 12. Let $T\to +\infty$, $H,U = o(T)$, $\sqrt{T} = o(U)$. Then the integral $I(n)$ can be represented in the form Proof. Put $\tau = \pi\sqrt{n}(\sqrt{t+U}-\sqrt{t})$ in the original integral. Then
$$
\begin{equation*}
\begin{gathered} \, \sqrt{t}\, dt=-\frac{1}{4\pi^3}\,\frac{1}{\tau^4n^{3/2}}(\pi^4n^2U^2-\tau^4) (\pi^2nU-\tau^2)\, d\tau, \\ \bigl|e^{2\pi i\sqrt{n}(\sqrt{t+U}-\sqrt{t})}-1\bigr|^2 =|e^{2i\tau}-1|^2=4\sin^2 \tau, \end{gathered}
\end{equation*}
\notag
$$
so that
$$
\begin{equation*}
I(n)=\frac{1}{(\pi n)^3}\int_{\sigma_n}^{\tau_n} (\pi^4n^2U^2-\tau^4)(\pi^2nU-\tau^2)\frac{\sin^2 \tau}{\tau^4}\, d\tau,
\end{equation*}
\notag
$$
where we are using the notation $\sigma_n = \pi\sqrt{n}(\sqrt{T+H+U}-\sqrt{T+H})$ along with $\tau_n$. Further rewriting $I(n)$, we have
$$
\begin{equation*}
\begin{aligned} \, I(n) &=\frac{1}{(\pi n)^3} \int_{\sigma_n}^{\tau_n} \bigl(\pi^{6}n^3U^3-\pi^4n^2U^2\tau^2-\pi^2nU\tau^4+\tau^{6}\bigr) \frac{\sin^2 \tau}{\tau^4}\,d\tau \\ &=(\pi U)^3\int_{\sigma_n}^{\tau_n}\frac{\sin^2 \tau}{\tau^4}\,d\tau-\frac{\pi U^2}{n}\int_{\sigma_n}^{\tau_n}\frac{\sin^2 \tau}{\tau^2}\,d\tau-\frac{U}{\pi n^2} \int_{\sigma_n}^{\tau_n}\sin^2 \tau\, d\tau \\ &\qquad+\frac{1}{(\pi n)^3}\int_{\sigma_n}^{\tau_n}\tau^2\sin^2 \tau\, d\tau=(\pi U)^3A(n)-B(n)-C(n)+D(n). \end{aligned}
\end{equation*}
\notag
$$
To estimate the quantities $\Sigma_{B}$, $\Sigma_{C}$ and $\Sigma_{D}$, we put $h_n = \tau_n-\sigma_n$ and notice that
$$
\begin{equation*}
\begin{gathered} \, \tau_n=\pi\sqrt{nT}\biggl(\biggl(1+\frac{U}{T}\biggr)^{1/2}-1\biggr)=\frac{\pi U}{2\sqrt{T}}\sqrt{n}\biggl(1-\frac{U}{4T}+O\biggl(\frac{U^2}{T^2}\biggr)\biggr) \asymp\frac{U\sqrt{n}}{\sqrt{T}}, \\ \sigma_n \asymp \tau_n \asymp \frac{U\sqrt{n}}{\sqrt{T}}. \end{gathered}
\end{equation*}
\notag
$$
Furthermore, put
$$
\begin{equation*}
\begin{gathered} \, h=\bigl(\sqrt{T+U}-\sqrt{T}\bigr)-\bigl(\sqrt{T+H+U}-\sqrt{T+H}\bigr), \\ f(v)=\sqrt{v},\qquad g(v)=f(v)-f(v+H). \end{gathered}
\end{equation*}
\notag
$$
Then a repeated use of the mean-value theorem yields that
$$
\begin{equation*}
\begin{aligned} \, h &=g(T+U)-g(T)=Ug'(t)+\frac{1}{2}U^2g''(T+\theta_1U) \\ &=U\bigl(f'(T)-f'(T+H)\bigr)+\frac{1}{2}U^2\bigl(f''(T+\theta_1U)-f''(T+\theta_1U+H)\bigr) \\ &=U\biggl(-Hf''(T)-\frac{1}{2}H^2f^{(3)}(T+\theta_2H)\biggr)-\frac{1}{2}HU^2f^{(3)}(T+\theta_1U +\theta_3H) \\ &=-HUf''(T)-\frac{1}{2}H^2Uf^{(3)}(T+\theta_2H)-\frac{1}{2}HU^2f^{(3)}(T+\theta_1U+\theta_3H) \\ &=\frac{HU}{4T^{3/2}}-\frac{3H^2U}{16T^{5/2}}\biggl(1+\frac{\theta_2H}{T}\biggr)^{-5/2} -\frac{3HU^2}{16T^{5/2}}\biggl(1+\frac{\theta_1U+\theta_3H}{T}\biggr)^{-5/2} \\ &=\frac{HU}{4T^{3/2}}\biggl(1-\frac{3(H+U)}{4T}+O\biggl(\frac{(H+U)^2}{T^2}\biggr)\biggr). \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
\begin{aligned} \, h_n &=\pi h\sqrt{n}=\frac{\pi}{4}\,\frac{HU}{T^{3/2}}\,\sqrt{n} \biggl(1-\frac{3(H+U)}{4T}+O\biggl(\frac{(H+U)^2}{T^2}\biggr)\biggr) \\ &=g_n\biggl(1-\frac{3(H+U)}{4T}+O\biggl(\frac{(H+U)^2}{T^2}\biggr)\biggr)=g_n+\varepsilon_n, \end{aligned}
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
g_n=\frac{\pi}{4}\,\frac{HU}{T^{3/2}}\,\sqrt{n}=\frac{\pi\varkappa}{4}\sqrt{n},\qquad \varepsilon_n\ll \frac{H+U}{T}g_n\ll \frac{(H+U)HU}{T^{5/2}}\sqrt{n}.
\end{equation*}
\notag
$$
Thus we obtain that
$$
\begin{equation}
\begin{gathered} \, B(n)\leqslant \frac{\pi U^2}{n}\int_{\sigma_n}^{\tau_n}\frac{d\tau}{\tau^2}\leqslant \frac{\pi U^2}{n}\,\frac{h_n}{\sigma_n^2}\ll \frac{HU}{\sqrt{T}}\,\frac{1}{n^{3/2}}, \\ \Sigma_{B}\ll \frac{HU}{\sqrt{T}}\ll H\sqrt{T}\,\frac{U}{T}; \end{gathered}
\end{equation}
\tag{16}
$$
$$
\begin{equation}
\begin{gathered} \, C(n)\leqslant \frac{U^2}{\pi n^2}\int_{\sigma_n}^{\tau_n}\, d\tau\leqslant \frac{Uh_n}{\pi n^2}\ll \frac{HU^2}{T^{3/2}}\,\frac{1}{n^{3/2}}, \\ \Sigma_{C}\ll \frac{HU^2}{T^{3/2}}\ll H\sqrt{T}\,\frac{U^2}{T^2}\ll H\sqrt{T}\,\frac{U}{T}. \end{gathered}
\end{equation}
\tag{17}
$$
Finally,
$$
\begin{equation}
\begin{gathered} \, D(n)\leqslant \frac{1}{(\pi n)^3}\int_{\sigma_n}^{\tau_n}\tau^2\, d\tau\leqslant \frac{\tau_n^2h_n}{(\pi n)^3}\ll \frac{HU^3}{T^{5/2}}\,\frac{1}{n^{3/2}}, \\ \Sigma_{D}\ll \frac{HU^3}{T^{5/2}}\ll H\sqrt{T}\,\frac{U^3}{T^3}\ll H\sqrt{T}\,\frac{U}{T}. \end{gathered}
\end{equation}
\tag{18}
$$
Rewriting the quantity $A(n)$, we have
$$
\begin{equation*}
A(n)=\frac{1}{2}\int_{\sigma_n}^{\tau_n}\frac{1-\cos 2\tau}{\tau^4}\,d\tau=\frac{1}{6} \biggl(\frac{1}{\sigma_n^3}-\frac{1}{\tau_n^3}\biggr)- \frac{1}{4}\biggl(\frac{\sin 2\tau_n}{\tau_n^4}-\frac{\sin 2\sigma_n} {\sigma_n^4}\biggr)-b(n),
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
b(n)=\int_{\sigma_n}^{\tau_n}\frac{\sin 2\tau}{\tau^5}\,d\tau.
\end{equation*}
\notag
$$
Clearly,
$$
\begin{equation*}
|b(n)|\leqslant \frac{h_n}{\sigma_n^5}\ll \frac{HT}{U^4}\,\frac{1}{n^2},
\end{equation*}
\notag
$$
so that
$$
\begin{equation}
U^3\Sigma_{b}\ll \frac{HT}{U}\ll H\sqrt{T}\,\frac{\sqrt{T}}{U}.
\end{equation}
\tag{19}
$$
Furthermore,
$$
\begin{equation*}
\frac{1}{4}\biggl(\frac{\sin 2\tau_n}{\tau_n^4}-\frac{\sin 2\sigma_n} {\sigma_n^4}\biggr)=\frac{1}{4\tau_n^4}\bigl(\sin 2\tau_n-\sin 2\sigma_n\bigr)+c(n),
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
c(n)=\frac{1}{4}\sin 2\sigma_n\biggl(\frac{1}{\tau_n^4}-\frac{1}{\sigma_n^4}\biggr).
\end{equation*}
\notag
$$
We have
$$
\begin{equation}
|c(n)|\leqslant \frac{1}{4}\biggl(\frac{1}{\sigma_n^4}-\frac{1}{\tau_n^4}\biggr) =\int_{\sigma_n}^{\tau_n}\frac{d\tau}{\tau^5}\leqslant\frac{h_n}{\sigma_n^5}, \qquad U^3\Sigma_{c}\ll \frac{HT}{U}\ll H\sqrt{T}\,\frac{\sqrt{T}}{U}.
\end{equation}
\tag{20}
$$
Also note that
$$
\begin{equation}
\begin{aligned} \, &\frac{1}{4\tau_n^4}\bigl(\sin 2\tau_n-\sin 2\sigma_n\bigr) =\frac{1}{4\tau_n^4}\bigl(\sin 2\tau_n-\sin2(\tau_n-g_n)\bigr)+d(n) \nonumber \\ &\qquad=\frac{1}{2\tau_n^4}\sin g_n\cos(2\tau_n-g_n)+d(n),\quad \text{where}\quad d(n)=\frac{\sin 2(\tau_n-g_n)-\sin 2\sigma_n}{4\tau_n^4}. \end{aligned}
\end{equation}
\tag{21}
$$
Then we see from the definition of $\varepsilon_n$ that
$$
\begin{equation*}
\begin{gathered} \, d(n)=\frac{\sin2(\tau_n-g_n)-\sin2(\tau_n-g_n-\varepsilon_n)}{4\tau_n^4} =\frac{\sin\varepsilon_n \cos(2(\tau_n-g_n)-\varepsilon_n)}{2\tau_n^4}, \\ |d(n)|\leqslant \frac{\varepsilon_n}{2\tau_n^4}\ll \frac{(H+U)HU}{T^{5/2}}\sqrt{n} \frac{T^2}{U^4n^2}\ll \frac{H(H+U)}{U^3\sqrt{T}}\,\frac{1}{n^{3/2}}. \end{gathered}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation}
U^3\Sigma_{d}\ll \frac{H(H+U)}{\sqrt{T}}\ll H\sqrt{T}\,\frac{H+U}{T}.
\end{equation}
\tag{22}
$$
Finally,
$$
\begin{equation*}
\frac{1}{6}\biggl(\frac{1}{\sigma_n^3}-\frac{1}{\tau_n^3}\biggr)=\frac{h_n}{2\tau_n^4}+e(n)= \frac{g_n}{2\tau_n^4}+e(n)+f(n),
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
e(n)=\frac{1}{6}\biggl(\frac{1}{\sigma_n^3}-\frac{1}{\tau_n^3}-\frac{3h_n}{\tau_n^4}\biggr),\qquad f(n)=\frac{\varepsilon_n}{2\tau_n^4}.
\end{equation*}
\notag
$$
Similarly to (22), we have Applying the mean-value theorem twice to $e(n)$ for some $\theta_1$, $\theta_2$ with $0\leqslant \theta_1, \theta_2\leqslant 1$, we have
$$
\begin{equation*}
\begin{aligned} \, e(n) &=\frac{1}{6}\biggl(\frac{1}{(\tau_n-h_n)^3} -\frac{1}{\tau_n^3}-\frac{3h_n}{\tau_n^4}\biggr)= \frac{1}{6}\biggl(\frac{3h_n}{(\tau_n-\theta_1h_n)^4}-\frac{3h_n}{\tau_n^4}\biggr) \\ &=\frac{h_n}{2}\biggl(\frac{1}{(\tau_n-\theta_1h_n)^4}-\frac{1}{\tau_n^4}\biggr)= \frac{\theta_1h_n^2}{2(\tau_n-\theta_2h_n)^5}\ll \frac{h_n^2}{\tau_n^5}\ll \frac{H^2}{U^3\sqrt{T}}\,\frac{1}{n^{3/2}}, \end{aligned}
\end{equation*}
\notag
$$
whence
$$
\begin{equation}
U^3\Sigma_{e}\ll \frac{H^2}{\sqrt{T}}\ll H\sqrt{T}\,\frac{H}{T}.
\end{equation}
\tag{24}
$$
Thus,
$$
\begin{equation*}
\begin{aligned} \, A(n) &=\frac{g_n}{2\tau_n^4}+e(n)+f(n)-\frac{1}{2\tau_n^4}\sin g_n \cos(2\tau_n-g_n)-b(n)-c(n)-d(n) \\ &=\frac{g_n}{2\tau_n^4}\biggl(1-\frac{\sin g_n}{g_n}\cos(2\tau_n-g_n)\biggr)+a(n), \end{aligned}
\end{equation*}
\notag
$$
where $a(n) = e(n)+f(n)-b(n)-c(n)-d(n)$. Using the inequalities (19), (20), (22)–(24), we find that
$$
\begin{equation}
U^3\Sigma_{a}\ll H\sqrt{T}\biggl(\frac{H+U}{T}+\frac{\sqrt{T}}{U}\biggr).
\end{equation}
\tag{25}
$$
Finally, noticing that
$$
\begin{equation*}
(\pi U)^3\frac{g_n}{2\tau_n^4} =\frac{HU^4}{8(nT)^{3/2}}\,\frac{1}{(\sqrt{T+U}-\sqrt{T})^4} =\frac{H\sqrt{T}}{8n^{3/2}}\biggl(\sqrt{1+\frac{U}{T}}+1\biggr)^4
\end{equation*}
\notag
$$
and putting
$$
\begin{equation*}
(\pi U)^3\frac{g_n}{2\tau_n^4}\biggl(1-\frac{\sin g_n}{g_n}\cos(2\tau_n-g_n)\biggr) =\frac{2H\sqrt{T}}{n^{3/2}}+k(n),
\end{equation*}
\notag
$$
we have
$$
\begin{equation}
\begin{gathered} \, |k(n)|\leqslant \frac{H\sqrt{T}}{8n^{3/2}}\biggl(\biggl(\sqrt{1+\frac{U}{T}}\biggr)^4-1\biggr)\biggl(1 -\frac{\sin g_n}{g_n} \cos(2\tau_n-g_n)\biggr) \ll \frac{HU}{\sqrt{T}}\,\frac{1}{n^{3/2}}, \\ \Sigma_{k}\ll \frac{HU}{\sqrt{T}}\ll H\sqrt{T}\,\frac{U}{T}. \end{gathered}
\end{equation}
\tag{26}
$$
Let $J(n)$ be the difference between $I(n)$ and the quantity
$$
\begin{equation*}
\frac{2H\sqrt{T}}{n^{3/2}}\biggl(1-\frac{\sin g_n}{g_n}\cos(2\tau_n-g_n)\biggr).
\end{equation*}
\notag
$$
By (16)–(18), (25), (26) we have Corollary 5. Suppose that $T\to +\infty$, $H,U\,{=}\,o(T)$, $\sqrt{T}\,{=}\,o(U)$, $\sqrt{T}(\log T)^2\,{=}\,o(H)$, $N=T(4U)^{-1}$. Then we have Corollary 6. Under the hypotheses of Corollary 5 we have Proof. By Lemma 9,
$$
\begin{equation*}
\begin{aligned} \, I(T+U,H) &=\frac{A}{2\pi^2}H\sqrt{T+U}\biggl(1+O\biggl(\frac{H}{T}\biggr) +O\biggl(\frac{\sqrt{T}}{H}\varpi(T)\biggr) +O\biggl(\frac{\varphi(T)}{\sqrt[4]{T}}\biggr)\biggr) \\ &=\frac{A}{2\pi^2}H\sqrt{T}\biggl(1+O\biggl(\frac{H+U}{T}\biggr) +O\biggl(\frac{\sqrt{T}}{H}\varpi(T)\biggr)+O\biggl(\frac{\varphi(T)}{\sqrt[4]{T}}\biggr)\biggr), \end{aligned}
\end{equation*}
\notag
$$
where $\varpi(T) = (\log{T})^{3/2}\log\log{T}$. Therefore,
$$
\begin{equation*}
I(T+U,H)+I(T,H)\,{=}\,\frac{A}{\pi^2}H\sqrt{T}\biggl(1+O\biggl(\frac{H+U}{T}\biggr) +O\biggl(\frac{\sqrt{T}}{H}\varpi(T)\biggr) +O\biggl(\frac{\varphi(T)}{\sqrt[4]{T}}\biggr)\biggr).
\end{equation*}
\notag
$$
Substituting this expression into the equality
$$
\begin{equation*}
2\mathcal{K}_{P}(T;H,U)=I(T+U,H)+I(T,H)-\mathcal{Q}_{P}(T;H,U)
\end{equation*}
\notag
$$
and using Lemma 12 along with the equality we arrive at the desired assertion. $\Box$ The following lemma enables us to simplify the formula in Lemma 12 in the case when the ratio $\varkappa = HU/T^{3/2}$ is small. Lemma 13. Suppose that $T\geqslant T_0$, $D = D(T)\leqslant T^{1/6}$ is a monotone unbounded function of arbitrarily slow growth as $T\to +\infty$ and Proof. We use the identity
$$
\begin{equation}
\begin{gathered} \, \frac{\sin g_n}{g_n}\cos(2\tau_n-g_n)=\cos 2\tau_n+a(n), \\ a(n)=\cos 2\tau_n\biggl(\frac{\sin 2g_n}{2g_n}-1\biggr)+\sin 2\tau_n\frac{\sin^2 g_n}{g_n}. \end{gathered}
\end{equation}
\tag{27}
$$
It is easy to see that $N\geqslant T/(TD^{-2}) = D^2$. If $1\leqslant n\leqslant D^2$, then
$$
\begin{equation*}
g_n=\frac{\pi}{4}\,\frac{HU}{T^{3/2}}\sqrt{n}\leqslant \frac{\pi}{4D} D=\frac{\pi}{4}.
\end{equation*}
\notag
$$
Noticing that the maximum value of the function
$$
\begin{equation*}
\frac{1}{x}\biggl(1-\frac{\sin 2x}{2x}+\frac{\sin^2 x}{x}\biggr)
\end{equation*}
\notag
$$
on the interval $0\leqslant x\leqslant \pi/4$ is attained at the right endpoint, we obtain the following inequalities for these $n$:
$$
\begin{equation*}
|a(n)|\leqslant 1-\frac{\sin 2g_n}{2g_n}+\frac{\sin^2 g_n}{g_n}\leqslant \frac{4g_n}{\pi}.
\end{equation*}
\notag
$$
But if $D^2<n\leqslant N$, then
$$
\begin{equation*}
|a(n)|\leqslant 1-\frac{\sin 2g_n}{2g_n}+\biggl|\frac{\sin g_n}{g_n}\biggr|\leqslant 3.
\end{equation*}
\notag
$$
Therefore, if we replace the product $(\sin g_n/g_n)\cos(2\tau_n-g_n)$ in the formula in Lemma 12 by the quantity $\cos 2\tau_n$, then the absolute value of the resulting error does not exceed
$$
\begin{equation*}
\begin{aligned} \, &\frac{H\sqrt{T}}{2\pi^2}\biggl(\sum_{n\leqslant D^2}\frac{r^2(n)}{n^{3/2}}\,\frac{4g_n}{\pi}+3\sum_{D^2<n\leqslant N}\frac{r^2(n)}{n^{3/2}}\biggr) \\ &\qquad\leqslant \frac{H\sqrt{T}}{2\pi^2}\biggl(\sum_{n\leqslant D^2}\frac{r^2(n)}{n^{3/2}}\,\frac{\sqrt{n}}{D} +3\sum_{n>D^2}\frac{r^2(n)}{n^{3/2}}\biggr) \\ &\qquad\ll H\sqrt{T}\biggl(\frac{1}{D}\sum_{n\leqslant D^2}\frac{r^2(n)}{n} +\sum_{n>D^2}\frac{r^2(n)}{n^{3/2}}\biggr)\ll H\sqrt{T}\,\frac{(\log D)^2}{D}. \quad\Box \end{aligned}
\end{equation*}
\notag
$$
§ 4. Lemmas on simultaneous approximationSuppose that $\nu\geqslant 2$ and let $1\leqslant n_1<n_2<\dots<n_s\leqslant \nu$ be arbitrary square-free numbers. In this section we make some assertions about simultaneous approximations of $\sqrt{n_1}, \sqrt{n_2},\dots,\sqrt{n_s}$. Lemma 14. Suppose that $\nu\geqslant 2$, $s\geqslant 2$, $1\leqslant n_1<\dots<n_s\leqslant \nu$ are distinct square-free numbers and $m_1,\dots, m_s$ are integers not all equal to zero. Put $Q = m_1\sqrt{n_1}+\dots+m_s\sqrt{n_s}$. Then Proof. We follow the arguments of Selberg in his unpublished manuscript [11] (see also Lemma 5 in [12]). Consider the product
$$
\begin{equation*}
P=\prod_{(c_2,\dots,c_s)}\bigl(m_1\sqrt{n_1}+c_2m_2\sqrt{n_2}+\dots +c_sm_s\sqrt{n_s}\,\bigr),
\end{equation*}
\notag
$$
where each $c_j$ assumes the values $\pm 1$ independently of the others. Since $\sqrt{n_1},\dots,\sqrt{n_s}$ are linearly independent over the field of rational numbers (see, for example, [13], Ch. VIII, § 6), all the factors in $P$ are non-zero. On the other hand, $P$ does not change if we replace $\sqrt{n_j}$ by $(-\sqrt{n_j})$. When $2\leqslant j\leqslant s$, this is clear. When $j = 1$, it follows from the equalities
$$
\begin{equation*}
\begin{aligned} \, &\prod_{(c_2,\dots,c_s)}\bigl(-m_1\sqrt{n_1}+c_2m_2\sqrt{m_2}+\dots+c_sm_s\sqrt{n_s}\,\bigr) \\ &\qquad=(-1)^{2^{s-1}}\prod_{(c_2,\dots,c_s)}\bigl(m_1\sqrt{n_1}-c_2m_2\sqrt{m_2} -\dots-c_sm_s\sqrt{n_s}\,\bigr) \\ &\qquad=\prod_{(c_2',\dots,c_s')}\bigl(m_1\sqrt{n_1}+c_2'm_2\sqrt{m_2} +\dots+c_s'm_s\sqrt{n_s}\,\bigr)=P \end{aligned}
\end{equation*}
\notag
$$
since the tuple $(c_2',\dots,c_s') = (-c_2,\dots,-c_s)$ and the tuple $(c_2,\dots,c_s)$ run over the same set. Therefore all the terms arising from multiplying out in $P$ contain only even powers of $\sqrt{n_j}$. It follows that $P$ is an integer. Hence $|P|\geqslant 1$. Moreover, for any $c_2,\dots, c_s$ we have
$$
\begin{equation*}
a |m_1\sqrt{n_1}+c_2m_2\sqrt{n_2}+\dots+c_sm_s\sqrt{n_s}\,|\leqslant (|m_1|+\dots+|m_s|)\sqrt{\nu},
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
\begin{aligned} \, 1 &\leqslant |P|\leqslant |Q|(|m_1|+\dots+|m_s|)^{2^{s-1}-1}(\sqrt{\nu})^{2^{s-1}-1} \\ &\leqslant |Q|\nu^{2^{s-2}-0.5}s^{2^{s-1}-2}\sum_{j=1}^{s}|m_j|^{2^{s-1}-1}\leqslant |Q|\nu^{2^{\nu}}\sum_{j=1}^{s}|m_j|^{2^{s-1}-1}.\qquad\Box \end{aligned}
\end{equation*}
\notag
$$
Lemma 15. Let $\nu\geqslant 2$ be an integer. We fix a constant $\varepsilon$, $0<\varepsilon<10^{-2}$, and take $c = 0$ or $c = 1/2$. Then there are positive numbers $h$ and $\lambda$ (depending only on $\nu$ and $\varepsilon$) such that every interval $(y,y+h)$ contains an interval $E$ of length $\lambda$ with $\|t\sqrt{n}-c\|<\varepsilon$ for all $t\in E$ and all square-free numbers $n$, $1\leqslant n\leqslant \nu$. Proof. Let $r$ be the number of square-free numbers not exceeding $\nu$. We put $\alpha = c-\varepsilon/2$, $\beta = c+\varepsilon/2$, $\Delta = \varepsilon/2$, $R = 2^{r-1}$ and use these parameters to construct a function $\psi(x)$ of ‘Vinogradov glass’ type as described in [14], Ch. 1, § 2. Then $\psi(x+1) = \psi(x)$ for all $x$ and, moreover,
$$
\begin{equation*}
\begin{aligned} \, \psi(x) = 1\quad&\text{for}\quad c-\frac{1}{4}\varepsilon\leqslant x\leqslant c+\frac{1}{4}\varepsilon, \\ \psi(x) = 0\quad&\text{for}\quad c+\frac{3}{4}\varepsilon\leqslant x\leqslant 1+c-\frac{3}{4}\varepsilon, \\ 0<\psi(x)<1\quad&\text{for}\quad c-\frac{3}{4}\varepsilon<x<c-\frac{1}{4}\varepsilon\quad \text{and}\quad c+\frac{1}{4}\varepsilon<x<c+\frac{3}{4}\varepsilon. \end{aligned}
\end{equation*}
\notag
$$
The Fourier series expansion of $\psi(x)$ is of the form
$$
\begin{equation}
\psi(x)=\varepsilon+\mathop{{\sum}'}_{m = -\infty}^{+\infty}c(m)e^{2\pi imx},
\end{equation}
\tag{28}
$$
where the prime means that $m\ne 0$ and the coefficients $c(m)$ satisfy the condition
$$
\begin{equation*}
|c(m)|\leqslant g(m)=\min\biggl(\varepsilon,\frac{1}{\pi|m|},\frac{1}{\pi|m|} \biggl(\frac{2R}{|m|\varepsilon}\biggr)^{R}\biggr).
\end{equation*}
\notag
$$
We define a function $\Psi(t)$ by the formula
$$
\begin{equation}
\Psi(t)=\mathop{{\prod}''}_{1\leqslant n\leqslant \nu}\psi(t\sqrt{n}),
\end{equation}
\tag{29}
$$
where the double prime means the product over all square-free numbers $n\leqslant \nu$. It is easy to see that if $\Psi(t)>0$ for some $t$, then we have $\psi(t\sqrt{n})>0$ (and, therefore, $\|t\sqrt{n}-c\|\leqslant ({3}/{4})\varepsilon<\varepsilon$) for all $n\leqslant \nu$ under consideration. Given any $y$ and some positive $h$, we consider the integral
$$
\begin{equation*}
I=I(\nu,\varepsilon; h,y)=\int_{y}^{y+h}\Psi(t)\, dt.
\end{equation*}
\notag
$$
Replacing each factor in (29) by the expression (28) and multiplying out, we obtain the following equalities, where $1=n_1<n_2<\dots<n_{r}$ are all square-free numbers not exceeding $\nu$:
$$
\begin{equation*}
\begin{aligned} \, &\Psi(t) =\prod_{j=1}^r\biggl(\varepsilon+\mathop{{\sum}'}_{m_j}c(m_j)e^{2\pi itm_j\sqrt{n_j}}\biggr)=\varepsilon^r+\varepsilon^{r-1}\sum_{1\leqslant j\leqslant r}\mathop{{\sum}'}_{m_j}c(m_j)e^{2\pi itm_j\sqrt{n_j}} \\ &+\varepsilon^{r-2}\sum_{1\leqslant j_1<j_2\leqslant r}\, \mathop{{\sum}'}_{m_{j_1}, m_{j_2}}c(m_{j_1})c(m_{j_2})e^{2\pi it(m_{j_1} \sqrt{n_{j_1}}+m_{j_2}\sqrt{n_{j_2}})} \\ &+\dots+\varepsilon^{r-s}\!\!\!\!\sum_{1\leqslant j_1<\dots<j_s\leqslant r}\,\mathop{{\sum}'}_{m_{j_1},\dots, m_{j_s}}\!\!\!\!c(m_{j_1})\cdots c(m_{j_s})e^{2\pi it(m_{j_1}\sqrt{n_{j_1}} +\dots+m_{j_s}\sqrt{n_{j_s}})}+\cdots\,. \end{aligned}
\end{equation*}
\notag
$$
Integrating the resulting equation with respect to $t$, we obtain
$$
\begin{equation*}
I=h\varepsilon^r+\varepsilon^{r-1}\sum_{\mathbf{n}_1}W(\mathbf{n}_1) +\varepsilon^{r-2}\sum_{\mathbf{n}_2}W(\mathbf{n}_2) +\dots+\varepsilon^{r-s}\sum_{\mathbf{n}_s}W(\mathbf{n}_s) + \dotsb,
\end{equation*}
\notag
$$
where the $\mathbf{n}_s$ are tuples of the form $(n_{j_1},\dots, n_{j_s})$ and
$$
\begin{equation*}
\begin{aligned} \, W(\mathbf{n}_s) &=\mathop{{\sum}'}_{m_{j_1},\dots, m_{j_s} = -\infty}^{+\infty}c(m_{j_1})\cdots c(m_{j_s})\int_{y}^{y+h}e^{2\pi itQ}\, dt, \\ Q &= m_{j_1}\sqrt{n_{j_1}}+\dots+m_{j_s}\sqrt{n_{j_s}}. \end{aligned}
\end{equation*}
\notag
$$
Proceeding to perform the estimates and using the inequalities in Lemma 14, we find that
$$
\begin{equation*}
\begin{aligned} \, |W(\mathbf{n}_s)| &\leqslant \frac{1}{\pi}\mathop{{\sum}'}_{m_{j_1},\dots, m_{j_s} = -\infty}^{+\infty}g(m_{j_1})\cdots g(m_{j_s})|Q|^{-1} \\ &\leqslant \frac{\nu^{2^{\nu}}}{\pi}\mathop{{\sum}'}_{m_{j_1},\dots, m_{j_s} = -\infty}^{+\infty}g(m_{j_1})\cdots g(m_{j_s}) \bigl(|m_{j_1}|^{2^{s-1}-1}+\dots+|m_{j_s}|^{2^{s-1}-1}\bigr) \\ &\leqslant \frac{2^{s}\nu^{2^{\nu}}}{\pi} \sum_{m_{j_1},\dots, m_{j_s} = 1}^{+\infty}g(m_{j_1})\cdots g(m_{j_s}) \bigl(m_{j_1}^{2^{s-1}-1}+\dots+m_{j_s}^{2^{s-1}-1}\bigr) \\ &\leqslant \frac{s 2^{s}\nu^{2^{\nu}}}{\pi}\sum_{m_{j_1},\dots, m_{j_s} = 1}^{+\infty}g(m_{j_1})\cdots g(m_{j_s}) m_{j_1}^{2^{s-1}-1}. \end{aligned}
\end{equation*}
\notag
$$
Since
$$
\begin{equation*}
g(m)\leqslant \frac{1}{\pi m}\biggl(\frac{2R}{m\varepsilon}\biggr)^{R}\quad\text{for}\quad m\geqslant 1,
\end{equation*}
\notag
$$
we have
$$
\begin{equation*}
|W(\mathbf{n}_s)|\leqslant \frac{s}{\pi}2^{s}\nu^{2^{\nu}}\biggl(\frac{2R}{\varepsilon}\biggr)^{sR} \frac{1}{\pi^{s}}\sum_{m_1=1}^{+\infty}\frac{m_1^{2^{s-1}-1}}{m_1^{R+1}} \biggl(\sum_{m=1}^{+\infty}\frac{1}{m^{R+1}}\biggr)^{s-1}.
\end{equation*}
\notag
$$
Since $R = 2^{r-1}\geqslant 2^{s-1}$, we have $(R+1)-(2^{s-1}-1)\geqslant 2$ and
$$
\begin{equation*}
\sum_{m=1}^{+\infty}\frac{m^{2^{s-1}-1}}{m^{R+1}}\leqslant \sum_{m=1}^{+\infty}\frac{1}{m^2}=\frac{\pi^2}{6}.
\end{equation*}
\notag
$$
Furthermore,
$$
\begin{equation*}
\begin{aligned} \, \sum_{m=1}^{+\infty}\frac{1}{m^{R+1}} &\leqslant 1+\int_1^{+\infty}\frac{du}{u^{R+1}}=1+\frac{1}{R}, \\ \biggl(\sum_{m=1}^{+\infty}\frac{1}{m^{R+1}}\biggr)^{s-1} &\leqslant \biggl(1+\frac{1}{R}\biggr)^{s}\leqslant e^{s/R}\leqslant e^{r/R} \leqslant \sqrt{e}, \end{aligned}
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
|W(\mathbf{n}_s)|\leqslant \frac{s 2^{s}\sqrt{e}}{6\pi^{s-1}} \biggl(\frac{2R}{\varepsilon}\biggr)^{sR}\nu^{2^{\nu}}< \frac{7}{10}\biggl(\frac{2R}{\varepsilon}\biggr)^{sR}\nu^{2^{\nu}}.
\end{equation*}
\notag
$$
The total number of distinct tuples $\mathbf{n}_s$ is equal to $\binom{r}{s}$. Therefore,
$$
\begin{equation*}
\begin{aligned} \, |I-h\varepsilon^r| &\leqslant \frac{7}{10}\nu^{2^{\nu}}\sum_{s=1}^r\binom{r}{s}\varepsilon^{r-s} \biggl(\frac{2R}{\varepsilon}\biggr)^{sR} \\ &=\frac{7}{10}\nu^{2^{\nu}}\biggl(\biggl(\frac{2R}{\varepsilon}\biggr)^{R}+1\biggr)^r= \frac{7}{10}\nu^{2^{\nu}}\biggl(\frac{2R}{\varepsilon}\biggr)^{R} \biggl(1+\frac{\varepsilon^{R+1}}{(2R)^{R}}\biggr)^r. \end{aligned}
\end{equation*}
\notag
$$
The last factor does not exceed
$$
\begin{equation*}
\exp{\biggl(\frac{r\varepsilon^{R+1}}{(2R)^{R}}\biggr)}=\exp{\biggl(\varepsilon r\biggl(\frac{\varepsilon}{2R}\biggr)^{R}\biggr)}\leqslant \exp{\biggl(\varepsilon\biggl(\frac{\varepsilon r^{1/R}}{2R}\biggr)^{R}\biggr)}<\exp{\biggl(\varepsilon\biggl(\frac{\varepsilon}{R}\biggr)^{R}\biggr)}<\frac{10}{7},
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
|I-h\varepsilon^r|<\nu^{2^{\nu}}\biggl(\frac{2R}{\varepsilon}\biggr)^{Rr}.
\end{equation*}
\notag
$$
Putting
$$
\begin{equation*}
h=\frac{2}{\varepsilon^r}\nu^{2^{\nu}}\biggl(\frac{3R}{r}\biggr)^{Rr},
\end{equation*}
\notag
$$
we find that
$$
\begin{equation*}
\begin{aligned} \, I&>h\varepsilon^r-\nu^{2^{\nu}}\biggl(\frac{2R}{\varepsilon}\biggr)^{Rr}=h\varepsilon^r \biggl(1-\frac{\nu^{2^{\nu}}}{\varepsilon^r}\,\frac{\varepsilon^r}{2\nu^{2^{\nu}}} \biggl(\frac{\varepsilon}{3R}\biggr)^{Rr}\biggl(\frac{2R}{\varepsilon}\biggr)^{Rr}\biggr) \\ &=h\varepsilon^r\biggl(1-\frac{1}{2}\biggl(\frac{2}{3}\biggr)^{Rr}\biggr) >\frac{2}{3}h\varepsilon^r. \end{aligned}
\end{equation*}
\notag
$$
Hence the interval $(y,y+h)$ contains a point $t_0$ such that $\Psi(t_0)>(2/3)\varepsilon^r$. Differentiation yields that
$$
\begin{equation*}
\Psi'(t)=\mathop{{\sum}''}_{n\leqslant \nu}\sqrt{n}\,\psi'(t\sqrt{n}) \mathop{{\prod}''}_{\substack{m\leqslant \nu \\ m\ne n}}\psi(t\sqrt{m}),
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
|\Psi'(t)|\leqslant \mathop{{\sum}''}_{n\leqslant \nu}\sqrt{n}\,|\psi'(t\sqrt{n})|.
\end{equation*}
\notag
$$
Differentiating the Fourier series for $\psi(x)$ term by term, we find that
$$
\begin{equation*}
\begin{gathered} \, \psi'(x)=2\pi i\mathop{{\sum}'}_{m=-\infty}^{+\infty}m c(m)e^{2\pi imx}, \\ |\psi'(x)|\leqslant 2\pi\mathop{{\sum}'}_{m=-\infty}^{+\infty}m g(m)\leqslant 2\mathop{{\sum}'}_{m=-\infty}^{+\infty} \biggl(\frac{2R}{|m|\varepsilon}\biggr)^{R} =4\biggl(\frac{2R}{\varepsilon}\biggr)^{R}\sum_{m=1}^{+\infty}\frac{1}{m^{R}}< \frac{9}{2}\biggl(\frac{2R}{\varepsilon}\biggr)^{R}. \end{gathered}
\end{equation*}
\notag
$$
We finally find that
$$
\begin{equation*}
|\Psi'(t)|\leqslant \frac{9}{2}\sqrt{\nu}\biggl(\frac{2R}{\varepsilon}\biggr)^{R} \frac{2\nu}{3}=3\nu^{3/2}\biggl(\frac{2R}{\varepsilon}\biggr)^{R}.
\end{equation*}
\notag
$$
Now put
$$
\begin{equation*}
\lambda=\frac{\nu^{-3/2}}{18}\biggl(\frac{\varepsilon}{2R}\biggr)^{R}\varepsilon^r.
\end{equation*}
\notag
$$
We define $E$ as whichever of the intervals $(t_0, t_0+\lambda)$, $(t_0,t_0-\lambda)$ is contained in $(y,y+h)$ (clearly, $\lambda < 1<h$). Then, for every $t\in E$, we have
$$
\begin{equation*}
\begin{gathered} \, \begin{aligned} \, |\Psi(t)-\Psi(t_0)|&=|\Psi'(\xi)(t-t_0)|\leqslant 3\nu^{3/2}\biggl(\frac{2R}{\varepsilon}\biggr)^{R}\lambda \\ &=3\nu^{3/2}\biggl(\frac{2R}{\varepsilon}\biggr)^{R}\frac{\nu^{-3/2}}{18} \biggl(\frac{\varepsilon}{2R}\biggr)^{R}\varepsilon^r:= \frac16 \varepsilon^r, \end{aligned} \\ \Psi(t)\geqslant \Psi(t_0)-\frac{1}{6}\,\varepsilon^r\geqslant \frac{2}{3}\,\varepsilon^r-\frac{1}{6}\,\varepsilon^r= \frac{1}{2}\,\varepsilon^r>0. \end{gathered}
\end{equation*}
\notag
$$
By what was said above, the desired assertion follows easily from this. $\Box$ Remark 1. This argument shows that the quantities $h$ and $\lambda$ can be taken in the form Therefore Lemma 15 remains valid if we take the values given above for the quantities $h=h(\nu,\varepsilon)$ and $\lambda = \lambda(\nu,\varepsilon)$. § 5. Lemmas on the value distribution of trigonometric seriesSuppose that $\tau$ is a real number and $\nu\geqslant 2$. Put
$$
\begin{equation*}
S(\tau)=\sum_{m=1}^{+\infty}\frac{r^2(m)}{m^{3/2}}\cos(2\pi\tau\sqrt{m}),\qquad S_{\nu}(\tau)=\sum_{m\leqslant \nu}\frac{r^2(m)}{m^{3/2}}\cos(2\pi\tau\sqrt{m}).
\end{equation*}
\notag
$$
In view of the equality
$$
\begin{equation*}
S_{\nu}(\tau)=S(\tau)+O\biggl(\frac{\log{\nu}}{\sqrt{\nu}}\biggr)
\end{equation*}
\notag
$$
and since the quantity $S_{\nu}(\tau)$ (when $\nu = N$) occurs in the expression for $k_{P}$ (see Lemma 13), it is natural to ask about the image of the function $S(\tau)$. Since
$$
\begin{equation*}
|S(\tau)|\leqslant S(0)=A=\sum_{m=1}^{+\infty}\frac{r^2(m)}{m^{3/2}},
\end{equation*}
\notag
$$
we have and the upper bound is clearly attainable. However, it turns out that the values of $S(\tau)$ lie in a narrower range: $-3A/4$ is a lower bound that cannot be replaced by a larger one. Our aim in this section is to prove these assertions. Lemma 16. Let $\varepsilon$ be sufficiently small, $0<\varepsilon<10^{-2}$. Suppose that $\nu\geqslant e^{6}$ and let $\tau$ be a real number such that the inequality holds for all square-free $n$, $1\leqslant n\leqslant \nu$. Then there are absolute constants $a,b>0$ such that Proof. Let $k$ be an integer such that
$$
\begin{equation}
3\leqslant k<\min{\biggl(\frac{1}{4\varepsilon},\sqrt{\nu}\biggr)}.
\end{equation}
\tag{30}
$$
Since every integer $m\geqslant 1$ can uniquely be represented in the form $q^2n$, where $n$ is square-free, we find that
$$
\begin{equation*}
\begin{aligned} \, S_{\nu}(\tau) &=\sum_{m\leqslant \nu}\frac{r^2(m)}{m^{3/2}} \cos(2\pi\tau\sqrt{m})=\sum_{q^2n\leqslant\nu}\frac{\mu^2(n)r^2(q^2n)}{(q^2n)^{3/2}} \cos(2\pi\tau q\sqrt{n}) \\ &=\sum_{q\leqslant \sqrt{\nu}}\frac{1}{q^3}\sum_{n\leqslant \nu/q^2} \frac{\mu^2(n)r^2(q^2n)}{n^{3/2}}\cos(2\pi\tau q\sqrt{n}) \\ &=\biggl(\sum_{q\leqslant k}+\sum_{k<q\leqslant \sqrt{\nu}}\,\biggr)\frac{1}{q^3}\sum_{n\leqslant \nu/q^2} \frac{\mu^2(n)r^2(q^2n)}{n^{3/2}}\cos(2\pi\tau q\sqrt{n}). \end{aligned}
\end{equation*}
\notag
$$
We claim that $h(q^2n)\leqslant h(q^2)h(n)$ for all square-free $n$. Indeed, if $n$ has a prime divisor $p$ of the form $4\ell+3$, then only odd powers of $p$ occur in the expansions of $q^2n$ and $n$ and, therefore, $h(q^2n) = h(n) = 0$. Otherwise we can represent $q$ and $n$ in the form $q = 2^{\alpha}q_1q_2$ and $n = 2^{\beta}n_1$, where $\alpha,\beta\geqslant 0$ and all the prime divisors of $q_1$, $n_1$ (resp. $q_2$) are of the form $4\ell + 1$ (resp. $4\ell+3$). Noticing that $h(q_1^2n_1) = \tau(q_1^2n_1)$ and using the inequality $\tau(uv)\leqslant \tau(u)\tau(v)$, which holds for all integers $u,v\geqslant 1$, we have
$$
\begin{equation*}
\begin{aligned} \, h(q^2n) &=h(2^{2\alpha+\beta}q_1^2q_2^2n_1)=h(q_1^2n_1)=\tau(q_1^2n_1) \\ &\leqslant \tau(q_1^2)\tau(n_1)= h(q_1^2)h(n_1) =h(2^{2\alpha}q_1^2q_2^2)h(2^{\beta}n_1)=h(q^2)h(n). \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
r(q^2n)=4h(q^2n)\leqslant 4h(q^2)h(n)=\frac{1}{4}r(q^2)r(n).
\end{equation*}
\notag
$$
Using Lemma 5, we conclude that the modulus of the contribution of $q>k$ to $S_{\nu}(\tau)$ does not exceed
$$
\begin{equation*}
\begin{aligned} \, &\frac{1}{16}\sum_{q>k}\frac{1}{q^3}\sum_{n\leqslant \nu/q^2} \frac{r^2(q^2)r^2(n)\mu^2(n)}{n^{3/2}} \leqslant \frac{1}{16}\sum_{q>k}\frac{r^2(q^2)}{q^3} \sum_{n=1}^{+\infty}\frac{\mu^2(n)r^2(n)}{n^{3/2}} \\ &\qquad\ll \sum_{q>k}\frac{r^2(q^2)}{q^3}\leqslant c_1\frac{(\log{k})^4}{k^2}. \end{aligned}
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
S_{\nu}(\tau)=S_1+\theta_1c_1\frac{(\log{k})^4}{k^2},\qquad S_1=\sum_{q\leqslant k}\frac{1}{q^3}\sum_{n\leqslant \nu/q^2}\frac{\mu^2(n)r^2(q^2n)}{n^{3/2}}\cos(2\pi q\sqrt{n}).
\end{equation*}
\notag
$$
By hypothesis, for every square-free $n\leqslant\nu$ there is an integer $k_n$ such that
$$
\begin{equation*}
\tau\sqrt{n}=k_n+\theta(n)\frac{\varepsilon}{2\pi},\qquad |\theta(n)|\leqslant 1.
\end{equation*}
\notag
$$
Hence, for every integer $q$, $1\leqslant q\leqslant k \leqslant (4\varepsilon)^{-1}$, we have
$$
\begin{equation*}
\begin{gathered} \, 2\pi\tau q\sqrt{n}=2\pi qk_n+\theta(n)q\varepsilon, \\ \cos(2\pi\tau q\sqrt{n}) =\cos(\theta(n)q\varepsilon) =1-\frac{1}{2}\theta'(n,q)(q\varepsilon)^2,\qquad |\theta'(n,q)|\leqslant 1. \end{gathered}
\end{equation*}
\notag
$$
We conclude that
$$
\begin{equation*}
S_1=\sum_{q\leqslant k}\frac{1}{q^3}\sum_{n\leqslant \nu/q^2}\frac{\mu^2(n)r^2(q^2n)}{n^{3/2}} \biggl(1-\frac{1}{2}\theta'(n,q)(q\varepsilon)^2\biggr).
\end{equation*}
\notag
$$
The modulus of the contribution of the last summand in brackets does not exceed
$$
\begin{equation*}
\frac{\varepsilon^2}{2\cdot 16}\sum_{q\leqslant k}\frac{r^2(q^2)}{q} \sum_{n=1}^{+\infty}\frac{\mu^2(n)r^2(n)}{n^{3/2}}\ll \varepsilon^2\sum_{q\leqslant k}\frac{r^2(q^2)}{q}\leqslant c_2\varepsilon^2(\log{k})^5,
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
S_1=S_2+\theta_2c_2\varepsilon^2(\log{k})^5,\qquad S_2=\sum_{q\leqslant k}\frac{1}{q^3}\sum_{n\leqslant \nu/q^2}\frac{\mu^2(n)r^2(q^2n)}{n^{3/2}}.
\end{equation*}
\notag
$$
If we replace the sum over $n\leqslant \nu/q^2$ by the infinite sum, then the resulting error does not exceed
$$
\begin{equation*}
\begin{aligned} \, &\sum_{q\leqslant k}\frac{1}{q^3}\sum_{n>\nu/q^2}\frac{\mu^2(n)r^2(q^2n)}{n^{3/2}}\leqslant \frac{1}{16} \sum_{q\leqslant k}\frac{r^2(q^2)}{q^3}\sum_{n>\nu/q^2}\frac{\mu^2(n)r^2(n)}{n^{3/2}} \\ &\qquad\ll \sum_{q\leqslant k}\frac{r^2(q^2)}{q^3}\,\frac{\log{(\nu/q^2)}}{\sqrt{\nu/q^2}}\ll \frac{\log{\nu}}{\sqrt{\nu}} \sum_{q\leqslant k}\frac{r^2(q^2)}{q^2}\ll \frac{\log{\nu}}{\sqrt{\nu}}. \end{aligned}
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
S_2=\sum_{q\leqslant k}\frac{1}{q^3}\sum_{n=1}^{+\infty}\frac{\mu^2(n)r^2(q^2n)}{n^{3/2}} +\theta_3c_3\frac{\log{\nu}}{\sqrt{\nu}}= S_3+\theta_3c_3\frac{\log{\nu}}{\sqrt{\nu}}.
\end{equation*}
\notag
$$
Finally, if we replace the sum over $q\leqslant k$ by the infinite sum, the error does not exceed
$$
\begin{equation*}
\frac{1}{16}\sum_{q>k}\frac{r^2(q^2)}{q^3}\sum_{n=1}^{+\infty}\frac{\mu^2(n)r^2(n)}{n^{3/2}}\ll \sum_{q>k}\frac{r^2(q^2)}{q^3}\leqslant c_4\frac{(\log{k})^4}{k^2},
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
S_3=\sum_{q=1}^{+\infty}\sum_{n=1}^{+\infty}\frac{\mu^2(n)r^2(q^2n)}{(q^2n)^{3/2}}+ \theta_4c_4\frac{(\log{k})^4}{k^2}=A+\theta_4c_4\frac{(\log{k})^4}{k^2}.
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
S=A+\theta\biggl((c_1+c_4)\frac{(\log{k})^4}{k^2}+c_2\varepsilon^2(\log{k})^5 +c_3\frac{\log{\nu}}{\sqrt{\nu}}\biggr)=A+\theta R.
\end{equation*}
\notag
$$
If $(4\varepsilon)^{-1}\leqslant \sqrt{\nu}$, then we put $k = [(5\varepsilon)^{-1}]+1$ and obtain the following inequalities for $R$:
$$
\begin{equation*}
\begin{aligned} \, R &\leqslant (c_1+c_4)\frac{\bigl(\log{(1/\varepsilon)}\bigr)^4}{(5\varepsilon)^{-2}}+ c_2\varepsilon^2\biggl(\log\frac{1}{\varepsilon}\biggr)^5+c_3\frac{\log{\nu}}{\sqrt{\nu}} \\ &=\biggl(c_2+25(c_1+c_4)\biggl(\log{\frac{1}{\varepsilon}}\biggr)^{-1}\biggr)\varepsilon^2 \biggl(\log{\frac{1}{\varepsilon}}\biggr)^5+c_3\frac{\log{\nu}}{\sqrt{\nu}} \\ &\leqslant\frac{a\log{\nu}}{\sqrt{\nu}}+b\varepsilon^2\biggl(\log{\frac{1}{\varepsilon}}\biggr)^5, \qquad a=c_3,\quad b = c_2+\frac{25}{2\log{10}}(c_1+c_2). \end{aligned}
\end{equation*}
\notag
$$
But if $\sqrt{\nu}<(4\varepsilon)^{-1}$, then we put $k = [\sqrt{\nu}\,]$. Since $\nu\geqslant e^{6}$, we find that
$$
\begin{equation*}
\begin{aligned} \, R &\leqslant (c_1+c_4)\frac{(\log{\sqrt{\nu}})^4}{(0.5\sqrt{\nu})^2}+ c_2\varepsilon^2(0.5\log{\nu})^5+c_3\frac{\log{\nu}}{\sqrt{\nu}} \\ &=(c_1+c_4)\frac{(\log{\nu})^4}{4\nu}+c_3\frac{\log{\nu}}{\sqrt{\nu}} +c_2\varepsilon^2\biggl(\log{\frac{1}{\varepsilon}}\biggr)^5 \\ &=\frac{\log{\nu}}{\sqrt{\nu}}\biggl(c_3+(c_1+c_4)\frac{(\log{\nu})^3}{4\sqrt{\nu}}\biggr) +c_2\varepsilon^2\biggl(\log{\frac{1}{\varepsilon}}\biggr)^5 \\ &\leqslant\frac{a\log{\nu}}{\sqrt{\nu}}+b\varepsilon^2\biggl(\log{\frac{1}{\varepsilon}}\biggr)^5, \qquad a=c_3+\frac{1}{4}\biggl(\frac{6}{e}\biggr)^3(c_1+c_4),\quad b = c_2.\qquad\Box \end{aligned}
\end{equation*}
\notag
$$
Lemma 17. Let $\varepsilon$ be sufficiently small, $0<\varepsilon<10^{-2}$. Suppose that $\nu\geqslant e^{8}$ and let $\tau$ be a real number such that the inequality holds for all square-free $n$, $1\leqslant n\leqslant \nu$. Then there are absolute constants $a,b>0$ such that Proof. Choosing an integer $k$ satisfying the conditions (30), we find that
$$
\begin{equation*}
S_{\nu}(\tau)=S_1+\theta_1c_1\frac{(\log{k})^4}{k^2},\qquad S_1=\sum_{q\leqslant k}\frac{1}{q^3}\sum_{n\leqslant \nu/q^2}\frac{\mu^2(n)r^2(q^2n)}{n^{3/2}}\cos(2\pi\tau q\sqrt{n}).
\end{equation*}
\notag
$$
For every square-free $n$, $1\leqslant n\leqslant\nu$, there is an integer $k_n$ such that
$$
\begin{equation*}
\tau\sqrt{n}=k_n+\frac{1}{2}+\theta(n)\frac{\varepsilon}{2\pi},\qquad |\theta(n)|\leqslant 1.
\end{equation*}
\notag
$$
Therefore, for every integer $q$, $1\leqslant q\leqslant k$, we have
$$
\begin{equation*}
\begin{gathered} \, 2\pi\tau q\sqrt{n}=2\pi qk_n+\pi q+\theta(n)\varepsilon q, \\ \cos(2\pi\tau q\sqrt{n})=\cos(\pi q+\theta(n)\varepsilon q) =(-1)^{q}\cos(\theta(n)\varepsilon q)=(-1)^{q}-\frac{1}{2}\theta'(n,q)(\varepsilon q)^2, \end{gathered}
\end{equation*}
\notag
$$
where $|\theta'(n,q)|\leqslant 1$. Arguing as above, we successively find that
$$
\begin{equation*}
S_1=\sum_{q = 1}^{+\infty}\frac{(-1)^{q}}{q^3} \sum_{n=1}^{+\infty}\frac{\mu^2(n)r^2(q^2n)}{n^{3/2}}+ \theta_2c_2\varepsilon^2(\log{k})^5+\theta_3c_3\frac{\log{\nu}}{\sqrt{\nu}} +\theta_4c_4\frac{(\log{k})^4}{k^2}.
\end{equation*}
\notag
$$
Furthermore, we put
$$
\begin{equation*}
A_j=\sum_{q\equiv j\ (\operatorname{mod}2)}\frac{1}{q^3} \sum_{n=1}^{+\infty}\frac{\mu^2(n)r^2(q^2n)}{n^{3/2}},\qquad j = 0,1.
\end{equation*}
\notag
$$
Clearly, $A_0+A_1 = A$. On the other hand,
$$
\begin{equation*}
\begin{aligned} \, A_1 &=16\sum_{q\equiv 1\ (\operatorname{mod}2)}\frac{1}{q^3}\biggl(\sum_{n\equiv 1\ (\operatorname{mod}2)}\frac{\mu^2(n)h^2(q^2n)}{n^{3/2}}+\sum_{n\equiv 1\ (\operatorname{mod}2)}\frac{\mu^2(2n)h^2(2q^2n)}{(2n)^{3/2}}\biggr) \\ &=16\biggl(1+\frac{1}{2\sqrt{2}}\biggr)\sum_{q\equiv 1\ (\operatorname{mod}2)}\ \sum_{n\equiv 1\ (\operatorname{mod}2)} \frac{\mu^2(n)h^2(q^2n)}{(q^2n)^{3/2}} \\ &=\biggl(1+\frac{1}{2\sqrt{2}}\biggr)\sum_{m\equiv 1\ (\operatorname{mod}2)}\frac{r^2(m)}{m^{3/2}}. \end{aligned}
\end{equation*}
\notag
$$
Moreover,
$$
\begin{equation*}
\begin{aligned} \, A&=\sum_{m=1}^{+\infty}\frac{r^2(m)}{m^{3/2}}=16\prod_{p}\biggl(1+\frac{h^2(p)}{p^{3/2}}+ \frac{h^2(p^2)}{p^3}+\frac{h^2(p^3)}{p^{9/2}}+\cdots\biggr) \\ &=16\biggl(1+\frac{1}{2^{3/2}}+\frac{1}{2^3}+\frac{1}{2^{9/2}}+\cdots\biggr)\prod_{p\ne 2} \biggl(1+\frac{h^2(p)}{p^{3/2}}+\frac{h^2(p^2)}{p^3}+\frac{h^2(p^3)}{p^{9/2}}+\cdots\biggr) \\ &=\biggl(1-\frac{1}{2\sqrt{2}}\biggr)^{-1}\sum_{m\equiv 1\ (\operatorname{mod}2)} \frac{r^2(m)}{m^{3/2}}, \end{aligned}
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
\sum_{m\equiv 1\ (\operatorname{mod}2)} \frac{r^2(m)}{m^{3/2}} =\biggl(1-\frac{1}{2\sqrt{2}}\biggr)A,\qquad A_1=\biggl(1+\frac{1}{2\sqrt{2}}\biggr)\biggl(1-\frac{1}{2\sqrt{2}}\biggr)A=\frac{7}{8}A.
\end{equation*}
\notag
$$
Therefore,
$$
\begin{equation*}
A_0=A-A_1=\frac{1}{8}A,\qquad A_0-A_1=-\frac{7}{8}A+\frac{1}{8}A=-\frac{3}{4}A.
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
\begin{gathered} \, S_1=-\frac{3}{4}A+ \theta_2c_2\varepsilon^2(\log{k})^5+\theta_3c_3\frac{\log{\nu}}{\sqrt{\nu}}+ \theta_4c_4\frac{(\log{k})^4}{k^2}, \\ S_{\nu}(\tau)=-\frac{3}{4}A+\theta R,\qquad R\leqslant c_3\frac{\log{\nu}}{\sqrt{\nu}}+(c_1+c_4)\frac{(\log{k})^4}{k^2}+c_2\varepsilon^2(\log{k})^5. \end{gathered}
\end{equation*}
\notag
$$
Choosing $k$ as was done in the proof of Lemma 16, we arrive at the required assertion. $\Box$ Lemma 18. The minimum value of the function $\varphi(x)=\sum_{k = 0}^{+\infty}2^{-3k}\cos^2{(2^{k}x)}$ is equal to $1/7$. Proof. It suffices to minimize $\varphi(x)$ on the interval $0\leqslant x\leqslant \pi$. Put
$$
\begin{equation*}
\varphi_n(x)=\sum_{k = 0}^{n}\frac{1}{2^{3k}}\cos^2{(2^{k}x)},
\end{equation*}
\notag
$$
so that $\varphi(x)\geqslant \varphi_n(x)$ for all $x$. It is easy to see that the minimum of $\varphi_n(x)$ is attained at $x = \pi/2$ and Therefore, Letting $n$ tend to infinity, we conclude that $\varphi(x)\geqslant 1/7$ for all $x$. A direct verification shows that $\varphi(\pi/2) = 1/7$. The required assertion follows. $\Box$ Proof. We have
$$
\begin{equation*}
\begin{aligned} \, S(\tau)+A &=\sum_{m=1}^{+\infty}\frac{r^2(m)}{m^{3/2}}\bigl(1+\cos(2\pi\tau\sqrt{m})\bigr) \\ &=2\sum_{m=1}^{+\infty}\frac{r^2(m)}{m^{3/2}}\cos^2(\pi\tau\sqrt{m})= 32\sum_{m=1}^{+\infty}\frac{h^2(m)}{m^{3/2}}\cos^2(\pi\tau\sqrt{m}). \end{aligned}
\end{equation*}
\notag
$$
Divide the sum into two parts: over $m = 2^{2k}n$ and over $m = 2^{2k+1}n$, where $k\geqslant 0$, $n$ is odd. We obtain
$$
\begin{equation*}
\begin{aligned} \, &S(\tau)+A=32\mathop{{\sum}'}_{n=1}^{+\infty}\sum_{k=0}^{+\infty}\biggl( \frac{h^2(2^{2k}n)}{(2^{2k}n)^{3/2}}\cos^2(\pi\tau\, 2^{k}\sqrt{n})+ \frac{h^2(2^{2k+1}n)}{(2^{2k+1}n)^{3/2}}\cos^2(\pi\tau\,2^{k}\sqrt{2n})\biggr) \\ &\enspace =32\mathop{{\sum}'}_{n=1}^{+\infty}\sum_{k=0}^{+\infty}\biggl(\frac{h^2(n)}{2^{3k}n^{3/2}} \cos^2(\pi\tau\,2^{k}\sqrt{n})+ \frac{1}{2\sqrt{2}}\,\frac{h^2(n)}{2^{3k}n^{3/2}} \cos^2(\pi\tau\,2^{k}\sqrt{2n})\biggr) \\ &\enspace =32\mathop{{\sum}'}_{n=1}^{+\infty}\frac{h^2(n)}{n^{3/2}}\sum_{k=0}^{+\infty} \frac{1}{2^{3k}}\cos^2{(2^{k}\pi\tau\sqrt{n})}+\frac{32}{2\sqrt{2}} \mathop{{\sum}'}_{n=1}^{+\infty}\frac{h^2(n)}{n^{3/2}}\sum_{k=0}^{+\infty} \frac{1}{2^{3k}}\cos^2{(2^{k}\pi\tau\sqrt{2n})} \end{aligned}
\end{equation*}
\notag
$$
(the prime means summing over odd $n$). By Lemma 18, each inner sum is greater than or equal to $1/7$. Therefore,
$$
\begin{equation*}
S(\tau)+A\geqslant \frac{32}{7}\biggl(1+\frac{1}{2\sqrt{2}}\biggr)\mathop{{\sum}'}_{n=1}^{+\infty} \frac{h^2(n)}{n^{3/2}}.
\end{equation*}
\notag
$$
Noticing that
$$
\begin{equation*}
\begin{aligned} \, &16\mathop{{\sum}'}_{n=1}^{+\infty}\frac{h^2(n)}{n^{3/2}} =16\prod_{p\ne 2}\biggl(1+\frac{h^2(p)}{p^{3/2}}+\frac{h^2(p^2)}{p^3}+\cdots\biggr) \\ &\qquad =16\biggl(1+\frac{1}{2^{3/2}}+\frac{1}{2^3}+\frac{1}{2^{9/2}}+\cdots\biggr)^{-1} \prod_{p}\biggl(1+\frac{h^2(p)}{p^{3/2}}+\frac{h^2(p^2)}{p^3}+\cdots\biggr) \\ &\qquad =16\biggl(1-\frac{1}{2\sqrt{2}}\biggr)\sum_{m=1}^{+\infty}\frac{h^2(m)}{m^{3/2}}= \biggl(1-\frac{1}{2\sqrt{2}}\biggr)A, \end{aligned}
\end{equation*}
\notag
$$
we find that § 6. Proofs of Theorems 1–6 Proof of Theorem 1. We conclude from Corollary 2 of Lemma 12 that
$$
\begin{equation*}
|k_{P}(T;H,U)|\leqslant \sum_{n\leqslant N}\frac{r^2(n)}{n^{3/2}} \biggl|\frac{\sin g_n}{g_n}\biggr|+o(1).
\end{equation*}
\notag
$$
First assume that $\varepsilon_0\geqslant 1$. Then
$$
\begin{equation*}
\biggl|\frac{\sin g_n}{g_n}\biggr|\leqslant \frac{1}{g_n}=\frac{4}{\pi\varkappa\sqrt{n}}\leqslant \frac{4}{\pi\varepsilon_0}\,\frac{1}{\sqrt{n}},
\end{equation*}
\notag
$$
whence
$$
\begin{equation}
|k_{P}(T;H,U)|\leqslant \frac{4}{\pi\varepsilon_0}\sum_{n\leqslant N}\frac{r^2(n)}{n^2}+o(1)\leqslant \frac{4B}{\pi\varepsilon^0}+o(1).
\end{equation}
\tag{31}
$$
Now assume that $0<\varepsilon_0<1$. Since $g_1\geqslant \pi\varepsilon_0/4$, we have
$$
\begin{equation*}
\biggl|\frac{\sin g_n}{g_n}\biggr|\leqslant \frac{\sin (\pi\varepsilon_0/4)}{\pi\varepsilon_0/4},
\end{equation*}
\notag
$$
for every $n\geqslant 1$. Hence,
$$
\begin{equation}
|k_{P}(T;H,U)|\leqslant \frac{\sin (\pi\varepsilon_0/4)}{\pi\varepsilon_0/4} \sum_{n=1}^{+\infty}\frac{r^2(n)}{n^{3/2}}+o(1) =A\,\frac{\sin(\pi\varepsilon_0/4)}{\pi\varepsilon_0/4}+o(1).
\end{equation}
\tag{32}
$$
Since
$$
\begin{equation*}
\frac{\sin x}{x}\leqslant 1-\frac{x^2}{8}\quad\text{for}\quad 0\leqslant x\leqslant \frac{\pi}{2},
\end{equation*}
\notag
$$
we conclude from (32) that
$$
\begin{equation}
|k_{P}(T;H,U)|\leqslant A\biggl(1-\frac{1}{8}\biggl(\frac{\pi\varepsilon_0}{4}\biggr)^2\biggr)+o(1).
\end{equation}
\tag{33}
$$
Theorem 1 follows from (31) and (33). $\Box$ Proof of Theorem 2. Take an arbitrary value of $\varkappa$, $D^{-1}\leqslant \varkappa\leqslant \varepsilon_0$, and put $\nu = [(c_0/\varkappa)^2]$, where the absolute constant is chosen in such a way that $\nu\geqslant 44$. Then, for every $U$ satisfying the conditions $\sqrt{T}(\log T)^3\leqslant U\leqslant Te^{-2\sqrt{\log{T}}}$ and $N = T/(4U)$, we have
$$
\begin{equation}
44\leqslant \nu \leqslant \biggl(\frac{c_0}{\varkappa}\biggr)^2\leqslant (c_0D)^2=\bigl(c_0\varepsilon_1\sqrt{\log\log{T}}\bigr)^2=(c_0\varepsilon_1)^2\log\log T <N.
\end{equation}
\tag{34}
$$
Then
$$
\begin{equation*}
\biggl|\sum_{\nu<n\leqslant N}\frac{r^2(n)}{n^{3/2}}\,\frac{\sin g_n}{g_n}\cos(2\tau_n-g_n)\biggr|\leqslant \sum_{\nu>n}\frac{r^2(n)}{n^{3/2}}\ll \frac{\log{\nu}}{\sqrt{\nu}}\leqslant c_1\varkappa\log\frac{c_0}{\varkappa},
\end{equation*}
\notag
$$
where $c_1>0$ is an absolute constant. We now conclude from Corollary 2 of Lemma 12 that, for an arbitrary $H$ with $\sqrt{T}(\log T)^3\leqslant H\leqslant T(\log{T})^{-1}$,
$$
\begin{equation}
k_{P}(T;H,U)=\sum_{n\leqslant\nu}\frac{r^2(n)}{n^{3/2}}\,\frac{\sin g_n}{g_n} \cos(2\tau_n-g_n)+ \theta_1c_1\varkappa\log{\frac{c_0}{\varkappa}}+O\biggl(\frac{1}{\log{T}}\biggr).
\end{equation}
\tag{35}
$$
Again using the identity (27) and the inequality
$$
\begin{equation*}
1-\frac{\sin 2x}{2x}+\frac{\sin^2 x}{x}\leqslant 1,\qquad 0\leqslant x\leqslant \frac{\pi}{4},
\end{equation*}
\notag
$$
we conclude that if the factor $(\sin g_n/g_n)\cos(2\tau_n-g_n)$ is replaced by $\cos 2\tau_n$ in (35), then the modulus of resulting error does not exceed
$$
\begin{equation*}
\sum_{n\leqslant\nu}\frac{r^2(n)}{n^{3/2}}\biggl(1-\frac{\sin 2g_n}{2g_n}+\frac{\sin^2 g_n}{g_n}\biggr)\leqslant \sum_{n\leqslant\nu}\frac{r^2(n)}{n^{3/2}}\,\frac{4g_n}{\pi} =\varkappa\sum_{n\leqslant\nu}\frac{r^2(n)}{n}\leqslant c_2\varkappa\biggl(\log{\frac{c_0}{\varkappa}}\biggr)^2,
\end{equation*}
\notag
$$
where $c_2>0$ is an absolute constant. Thus,
$$
\begin{equation*}
k_{P}(T;H,U)=\sum_{n\leqslant\nu}\frac{r^2(n)}{n^{3/2}} \cos 2\tau_n +\theta_1c_1\varkappa\log{\frac{c_0}{\varkappa}} +\theta_2c_2\varkappa\biggl(\log{\frac{c_0}{\kappa}}\biggr)^2+O\biggl(\frac{1}{\log{T}}\biggr).
\end{equation*}
\notag
$$
Given any numbers $\varepsilon_0$ and $\nu$, Lemma 15 and the subsequent remark enable us to define a quantity $h = h(\nu,\varepsilon_0)<(4\pi\nu\varepsilon_0^{-1})^{2^{\nu}}$ in such a way that the interval $(y,y+h)$ contains values of $\tau$ satisfying the inequalities $\|\tau\sqrt{n}\|<\varepsilon_0/(2\pi)$ for all square-free $n$, $1\leqslant n\leqslant\nu$. If $\varepsilon_0$ is sufficiently large, then
$$
\begin{equation*}
\varkappa^2\leqslant \varepsilon_0^2\leqslant \frac{1}{3}c_0^2\varepsilon_0,\qquad \frac{4\pi}{\varepsilon_0}\leqslant \frac{2}{3}\biggl(\frac{c_0}{\varkappa}\biggr)^2<\nu.
\end{equation*}
\notag
$$
Therefore, using (34), we find that
$$
\begin{equation*}
h<\nu^{\,2\cdot 2^{\nu}}<\exp (e^{\nu})\leqslant \exp{(e^{(c_0D)^2})} =\exp (e^{(c_0\varepsilon_1)^2\log\log{T}}) <e^{\sqrt[3]{\log{T}}}.
\end{equation*}
\notag
$$
Taking $y = (\log T)^3$, we obtain $\tau\leqslant y+h<2e^{\sqrt[3]{\log{T}}}$. Putting $U_1 = \tau(2\sqrt{T}+\tau)$, we now have
$$
\begin{equation*}
\begin{gathered} \, 2\sqrt{T}(\log T)^3<U_1<5\sqrt{T}e^{\sqrt[3]{\log{T}}}<\sqrt{T}e^{\sqrt{\log{T}}}, \\ \tau_n=\pi\sqrt{n}\bigl(\sqrt{T+U_1}-\sqrt{T}\bigr)=\pi\tau\sqrt{n}. \end{gathered}
\end{equation*}
\notag
$$
By Lemma 16, we have the following equality in the notation of § 5:
$$
\begin{equation*}
\sum_{n\leqslant\nu}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n=S_{\nu}(\tau)=A+\theta\biggl(a\frac{\log{\nu}}{\sqrt{\nu}}+ b\varepsilon_0^2\biggl(\log{\frac{1}{\varepsilon_0}}\biggr)^5\biggr),
\end{equation*}
\notag
$$
where $a,b>0$ are absolute constants. Putting $H_1=\varkappa\, T^{3/2}/U_1$, we find that
$$
\begin{equation*}
H_1\geqslant \frac{T^{3/2}}{DU_1}\geqslant \frac{T^{3/2}}{\sqrt{T}}\, \frac{e^{-\sqrt{\log{T}}}}{\varepsilon_1\sqrt{\log\log{T}}}>Te^{-2\sqrt{\log{T}}},\qquad H_1\leqslant \frac{\varepsilon_0T^{3/2}}{U_1}<T(\log{T})^{-3}.
\end{equation*}
\notag
$$
For this choice of $U_1$ and $H_1$ we have
$$
\begin{equation*}
\begin{aligned} \, &k_{P}(T;H_1,U_1) \\ &\qquad =A+\theta\biggl(a\frac{\log{\nu}}{\sqrt{\nu}}+b\varepsilon_0^2 \biggl(\log{\frac{1}{\varepsilon_0}}\biggr)^5+c_1\varkappa\log{\frac{c_0}{\varkappa}}+ c_2\varkappa\biggl(\log{\frac{c_0}{\kappa}}\biggr)^2\biggr)+O\biggl(\frac{1}{\log{T}}\biggr) \\ &\qquad =A+O\biggl(\varepsilon_0\log^2{\frac{1}{\varepsilon_0}}\biggr). \end{aligned}
\end{equation*}
\notag
$$
The second assertion of the theorem can be proved in a similar way. The only difference is that we choose $\tau$ in such a way that $\|\tau\sqrt{n}-0.5\|\leqslant \varepsilon_0/(2\pi)$ for all square-free $n\leqslant \nu$. $\Box$ Proof of Theorem 3. By Lemma 13, for every $U\in (V,2V)$ we have
$$
\begin{equation*}
k_{P}(T;H,U)=\sum_{n\leqslant N}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n +\frac{1}{2}\theta_1\varepsilon, \qquad N = \frac{T}{4U}.
\end{equation*}
\notag
$$
Put
$$
\begin{equation*}
\nu=\biggl[\biggl(\frac{c_0}{\varepsilon}\log{\frac{1}{\varepsilon}}\biggr)^2\,\biggr],
\end{equation*}
\notag
$$
where $c_0\geqslant 1$ is a sufficiently large absolute constant. Then
$$
\begin{equation*}
\biggl|\sum_{\nu<n\leqslant N}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n\biggr|\leqslant \sum_{n>\nu}\frac{r^2(n)}{n^{3/2}} \ll \frac{\log{\nu}}{\sqrt{\nu}}<\frac{\varepsilon}{4}.
\end{equation*}
\notag
$$
Again using the notation of § 5, we have
$$
\begin{equation*}
k_{P}(T;H,U)=S_{\nu}(\tau)+\frac{3}{4}\theta_2\varepsilon,\qquad \tau=\sqrt{T+U}-\sqrt{T}.
\end{equation*}
\notag
$$
We shall show that $S_{\nu}(\tau)$ has many zeros on the interval $(VT^{-0.5},2VT^{-0.5})$ and the distance between neighbouring zeros is sufficiently large. Narrow neighbourhoods of these zeros will be used in the construction of $\mathcal{E}$. Define the following functions $F_{k}(\tau)$:
$$
\begin{equation*}
\begin{gathered} \, F_0(\tau)=S_{\nu}(\tau),\qquad F_{k}(\tau)=\int_0^{\tau}F_{k-1}(t)\, dt+C_{k},\quad\text{where } C_{k}=0\text{ for odd }k, \\ C_{k}=\frac{(-1)^{\ell}}{(2\pi)^{2\ell}}\sum_{n\leqslant\nu}\frac{r^2(n)}{n^{\ell+3/2}} \quad\text{for}\quad k = 2\ell,\quad \ell\geqslant 1. \end{gathered}
\end{equation*}
\notag
$$
It is easily verifiable that $S_{\nu}(\tau) = F_{k}^{(k)}(\tau)$ and
$$
\begin{equation}
F_{k}(\tau)=\frac{1}{(2\pi)^{k}}\sum_{n\leqslant\nu}\frac{r^2(n)}{n^{(k+3)/2}}\, \phi_{k}(2\pi\tau\sqrt{n}),\ \text{where }\ \phi_{k}(t)= \begin{cases} \cos t, &k\equiv 0\ (\operatorname{mod}4), \\ \sin t, &k\equiv 1\ (\operatorname{mod}4), \\ -\cos t, &k\equiv 2\ (\operatorname{mod}4), \\ -\sin t, & k\equiv 3\ (\operatorname{mod}4). \end{cases}
\end{equation}
\tag{36}
$$
The first three terms of the series (36) are equal to
$$
\begin{equation*}
16\phi_{k}(2\pi\tau),\quad \frac{16}{2^{(k+3)/2}}\,\phi_{k}(2\pi\tau\sqrt{2}),\quad 0
\end{equation*}
\notag
$$
respectively. We conclude from Lemma 5 that the modulus of the sum of the other terms does not exceed
$$
\begin{equation*}
\sum_{4\leqslant n\leqslant\nu}\frac{r^2(n)}{n^{(k+3)/2}}\leqslant 16c_3 \frac{(k+3)}{2^{k+3}}.
\end{equation*}
\notag
$$
Thus,
$$
\begin{equation*}
F_{k}(\tau)=\frac{16}{(2\pi)^{k}}\biggl(\phi_{k}(2\pi\tau) +\frac{\phi_{k}(2\pi\tau\sqrt{2})}{2^{(k+3)/2}}+\theta \frac{c_3(k+3)}{2^{k+3}}\biggr).
\end{equation*}
\notag
$$
Choosing $k\equiv 1\pmod{4}$ so large that the modulus of the sum of the last two summands does not exceed $1/2$, we obtain
$$
\begin{equation*}
F_{k}(\tau)=\frac{16}{(2\pi)^{k}}\bigl(\sin (2\pi\tau)+0.5\theta'\bigr).
\end{equation*}
\notag
$$
Hence, for any pair of points $\tau_s = s-0.25$, $\sigma_s = s+0.25$ (where $s$ is an integer), we have
$$
\begin{equation*}
F_{k}(\tau_s)=\frac{16}{(2\pi)^{k}}\bigl(-1+0.5\theta_1'\bigr)<0,\qquad F_{k}(\sigma_s)=\frac{16}{(2\pi)^{k}}\bigl(1+0.5\theta_2'\bigr)>0.
\end{equation*}
\notag
$$
It follows that $F_{k}(\tau)$ has a zero in every interval of the form $(s-1/4,s+1/4)$ and, therefore, it has at least $k+1$ distinct zeros in every interval of the form $(s-1/4,s+(k+1)+1/4)$. By Rolle’s theorem, $F_{k}'(\tau)$ has at least $k$ distinct zeros in such an interval, $F_{k}''(\tau)$ has at least $k-1$ distinct zeros, …, $F_{k}^{(r)}(\tau)$ has at least $k-r+1$ distinct zeros, …, and $F_{k}^{(k)}(\tau) = S_{\nu}(\tau)$ has at least one zero. Take $s_0 = [VT^{-0.5}]+2$, so that $s_0-1/4>VT^{-0.5}$. Let $\xi_1,\xi_2,\dots,\xi_{m}$ be the zeros of $S_{\nu}(\tau)$ in the intervals
$$
\begin{equation*}
\begin{aligned} \, &E_1=\biggl(s_0-\frac{1}{4},s_0+(k+1)+\frac{1}{4}\biggr), \\ &E_2=\biggl(s_0+(k+2)-\frac{1}{4},s_0+(2k+3)+\frac{1}{4}\biggr), \\ &\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots \\ &E_{m}=\biggl(s_0+(m-1)(k+2)-\frac{1}{4},s_0+m(k+2)-1+\frac{1}{4}\biggr), \end{aligned}
\end{equation*}
\notag
$$
where the integer $m$ is determined by the inequalities
$$
\begin{equation*}
s_0+m(k+2)-1+\frac{1}{4}\leqslant \frac{2V}{\sqrt{T}}<s_0+(m+1)(k+2)-1+\frac{1}{4}.
\end{equation*}
\notag
$$
Then it is clear that
$$
\begin{equation*}
\xi_{j+1}-\xi_j>\frac{1}{2}\quad\text{and}\quad m>\frac{1}{k+2}\biggl(\frac{V}{\sqrt{T}} +\biggl\{\frac{V}{\sqrt{T}}\biggr\}+\frac{3}{4}\biggr)>\frac{1}{k+2}\,\frac{V}{\sqrt{T}}.
\end{equation*}
\notag
$$
We also notice that
$$
\begin{equation*}
S_{\nu}'(\tau)=-2\pi\sum_{n\leqslant\nu}\frac{r^2(n)}{n}\sin (2\pi\tau\sqrt{n}),
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
|S_{\nu}'(\tau)|\leqslant 2\pi\sum_{n\leqslant\nu}\frac{r^2(n)}{n}\leqslant c_4(\log \nu)^2.
\end{equation*}
\notag
$$
Therefore, putting $\delta = \varepsilon(4c_4(\log \nu)^2)^{-1}$ and using the Lagrange finite-increments formula, we have
$$
\begin{equation*}
S_{\nu}(\tau)=S_{\nu}(\tau)-S_{\nu}(\xi_j)=(\tau-\xi_j)S_{\nu}'(\eta),\qquad |S_{\nu}(\tau)|<\delta|S_{\nu}'(\eta)|\leqslant \delta c_4(\log \nu)^2=\frac{\varepsilon}{4}
\end{equation*}
\notag
$$
for every $\tau\in (\xi_j-\delta,\xi_j+\delta)$. The intervals $(\xi_j-\delta,\xi_j+\delta)$, $1\leqslant j\leqslant m$, are disjoint and we have $|S_{\nu}(\tau)|<\varepsilon/4$ on their union. Put
$$
\begin{equation*}
\mathcal{E}_j=\bigl((\xi_j-\delta)(2\sqrt{T}+\xi_j-\delta),(\xi_j+\delta)(2\sqrt{T}+\xi_j+\delta)\bigr)
\end{equation*}
\notag
$$
and let $\mathcal{E}$ be the union of all $\mathcal{E}_j$, $1\leqslant j\leqslant m$. Then, for every $U\in \mathcal{E}$, we have It is also clear that as required. $\Box$ Proof of Theorem 4. By Lemma 13, for any $U$ satisfying the conditions $V<U<2V$ and $N = T/(4U)$, we have
$$
\begin{equation*}
k_{P}(H;T,U)=\sum_{n\leqslant N}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n+o(1).
\end{equation*}
\notag
$$
Putting
$$
\begin{equation*}
\nu=\biggl[\biggl(\frac{c_0}{\varepsilon}\log{\frac{1}{\varepsilon}}\biggr)^2\biggr]+1
\end{equation*}
\notag
$$
for sufficiently large $c_0$, we obtain
$$
\begin{equation*}
k_{P}(H;T,U)=\sum_{n\leqslant N}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n+\theta_1\,\frac{\varepsilon}{2}.
\end{equation*}
\notag
$$
Then, given $\nu$ and $\varepsilon$, we use Lemma 14 to define numbers $h<(4\pi\nu\varepsilon^{-1})^{2^{\nu}}$ and $\lambda\geqslant (\varepsilon/(2\pi))^{2^{\nu}}$ in such a way that every interval $(y,y+h)$ contains an interval $E$ of length $\lambda$ such that the following inequalities hold for every $\tau\in E$ and all square-free integers $n$, $1\leqslant n\leqslant \nu$: Divide the interval $(V T^{-0.5}, 2V T^{-0.5})$ into at least $h^{-1}[V T^{-0.5}]$ intervals of the form
$$
\begin{equation*}
\biggl(\frac{V}{\sqrt{T}}+(m-1)h,\frac{V}{\sqrt{T}}+mh\biggr],\qquad 1\leqslant m\leqslant \frac{1}{h}\biggl[\frac{V}{\sqrt{T}}\biggr].
\end{equation*}
\notag
$$
In each of them, we distinguish an interval $E_{m}$ of length $\lambda$ such that (37) holds for all $\tau\in E_{m}$ and all square-free $n$, $1\leqslant n\leqslant \nu$. Finally, for every $\tau\in E_{m}$ we define a number $U = \tau(2\sqrt{T}+\tau)$. Then for every interval $E_{m}$ we have the corresponding interval $\mathcal{E}_{m}$ of values of $U$ and its length is greater than or equal to $2|E_{m}|\sqrt{T}\geqslant 2\lambda\sqrt{T}$. Take the union of all the intervals $\mathcal{E}_{m}$ for $\mathcal{E}$. Then
$$
\begin{equation*}
\operatorname{mes}(\mathcal{E})\geqslant 2\lambda\sqrt{T}\, \frac{1}{h}\biggl[\frac{V}{\sqrt{T}}\biggr] \geqslant \frac{\lambda}{h}\,V=c(\varepsilon)V.
\end{equation*}
\notag
$$
Suppose that $U\in \mathcal{E}$. Then
$$
\begin{equation*}
\sqrt{T+U}-\sqrt{T} = \tau,\qquad 2\tau_n=2\pi\sqrt{n}\bigl(\sqrt{T+U}-\sqrt{T}\bigr)=2\pi\tau\sqrt{n}
\end{equation*}
\notag
$$
and (37) holds for all square-free $n\leqslant\nu$ . Then Lemma 15 yields that
$$
\begin{equation*}
\biggl|\sum_{n\leqslant\nu}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n-A\biggr|\leqslant a\,\frac{\log{\nu}}{\sqrt{\nu}}+b\varepsilon^2\biggl(\log{\frac{1}{\varepsilon}}\biggr)^5,
\end{equation*}
\notag
$$
where $a$, $b$ are absolute constants. Assuming that the constant $c_0$ is sufficiently large and $\varepsilon$ is sufficiently small, we have
$$
\begin{equation*}
\begin{gathered} \, a\frac{\log{\nu}}{\sqrt{\nu}}\,{\leqslant}\, a\frac{2\log((c_0/\varepsilon)\log(1/\varepsilon))} {(c_0/\varepsilon)\log(1/\varepsilon)} \,{=}\, \frac{2a}{c_0}\,\varepsilon \frac{\log(1/\varepsilon)+\log\log(1/\varepsilon)+\log c_0}{\log(1/\varepsilon)} \,{\leqslant}\, \frac{3a}{c_0}\,\varepsilon\,{<}\,\frac{\varepsilon}{4}, \\ b\varepsilon^2\biggl(\log{\frac{1}{\varepsilon}}\biggr)^5=\varepsilon b\varepsilon\biggl(\log{\frac{1}{\varepsilon}}\biggr)^5<\frac{\varepsilon}{4}. \end{gathered}
\end{equation*}
\notag
$$
The first assertion of the theorem follows from these inequalities. The second can be proved in a similar way by using Lemma 16. $\Box$ Proof of Theorem 5. This follows immediately from the formula in Lemma 13, the equalities
$$
\begin{equation*}
\sum_{n\leqslant N}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n=\sum_{n\leqslant\nu}\frac{r^2(n)}{n^{3/2}}\, \cos(2\pi\tau\sqrt{n})+O\biggl(\frac{\log{\nu}}{\sqrt{\nu}}\biggr),\qquad \tau = \sqrt{T+U}-\sqrt{T}
\end{equation*}
\notag
$$
and Lemma 19. $\Box$ Proof of Theorem 6. Taking $D = (\log T)^2$ in Lemma 13 and noticing that
$$
\begin{equation*}
\frac{HU}{T^{3/2}}\leqslant \frac{T^{3/2-\delta}}{T^{3/2}}=T^{-\delta}\leqslant \frac{1}{D},
\end{equation*}
\notag
$$
we find that
$$
\begin{equation*}
k_{P}(T;H,U)=\sum_{n\leqslant N}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n +O\biggl(\frac{1}{\log{T}}\biggr).
\end{equation*}
\notag
$$
Let $c_0$ be a sufficiently small absolute constant. We put
$$
\begin{equation*}
\nu=\biggl[\frac{c_0\delta\log{T}}{\log\log{T}}\biggr],
\end{equation*}
\notag
$$
so that $1<\nu<N$. Then, by Lemma 5,
$$
\begin{equation*}
\biggl|\sum_{\nu<n\leqslant N}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n\biggr|\ll \frac{\log{\nu}}{\sqrt{\nu}}\leqslant c_1\frac{(\log\log{T})^{3/2}}{\sqrt{\delta\log{T}}}.
\end{equation*}
\notag
$$
By Dirichlet’s theorem on simultaneous approximation (see, for example, [15], Appendix, § 8, Theorem 4), for every $t_0\geqslant 1$ and any integer $q\geqslant 2$ there is a $\tau$ such that $t_0\leqslant \tau\leqslant t_0q^{\nu}$ and $\|\tau\sqrt{n}\|\leqslant q^{-1}$ for every $n$, $1\leqslant n\leqslant\nu$. Defining $U$ by the relation $\tau = \sqrt{T+U}-\sqrt{T}$ and putting $\tau_n = \pi\tau\sqrt{n}$, we have $\|\tau_n/\pi\|\leqslant q^{-1}$. Hence for every $n$, $1\leqslant n\leqslant\nu$, there is an integer $k_n$ such that
$$
\begin{equation*}
\tau_n=\pi k_n+\frac{\pi\theta_n}{q},\quad |\theta_n|\leqslant 1,\qquad \cos 2\tau_n =1-\theta_n'\frac{2\pi^2}{q^2}.
\end{equation*}
\notag
$$
It follows that
$$
\begin{equation*}
\begin{aligned} \, \sum_{n\leqslant \nu}\frac{r^2(n)}{n^{3/2}}\cos 2\tau_n &=\sum_{n\leqslant \nu} \frac{r^2(n)}{n^{3/2}}\biggl(1-\theta_n'\frac{2\pi^2}{q^2}\biggr) \\ &=\sum_{n\leqslant \nu}\frac{r^2(n)}{n^{3/2}} -\frac{2\pi^2\theta}{q^2}\sum_{n=1}^{+\infty}\frac{r^2(n)}{n^{3/2}}=A-\sum_{n> \nu} \frac{r^2(n)}{n^{3/2}}+O\biggl(\frac{1}{q^2}\biggr) \\ &=A+O\biggl(\frac{1}{q^2}\biggr)+O\biggl(\frac{(\log\log{T})^{3/2}}{\sqrt{\delta\log{T}}}\biggr), \end{aligned}
\end{equation*}
\notag
$$
whence
$$
\begin{equation*}
\begin{aligned} \, k_{P}(T;H,U) &=A+O\biggl(\frac{1}{q^2}\biggr) +O\biggl(\frac{(\log\log{T})^{3/2}}{\sqrt{\delta\log{T}}}\biggr) +O\biggl(\frac{1}{\log{T}}\biggr) \\ &=A+O\biggl(\frac{1}{q^2}+\frac{(\log\log{T})^{3/2}}{\sqrt{\delta\log{T}}}\biggr). \end{aligned}
\end{equation*}
\notag
$$
Put $t_0 = \log{T}$. Then, taking $q = [\log{T}]$, we have
$$
\begin{equation*}
\tau\leqslant t_0q^{\nu}\leqslant (\log{T})^{\nu+1}\leqslant \exp{\biggl(\frac{2c_0\delta\log{T}}{\log\log{T}} \log\log{T}\biggr)}\log{T}=T^{2c_0\delta}\log{T}\leqslant \frac{1}{3}T^{\delta}
\end{equation*}
\notag
$$
if $c_0 < 1/2$. But then and, finally, Corollary 5 follows from this formula and the definition of $\mathcal{Q}_{P}(T;H,U)$. Corollary 6 follows from the estimate in Corollary 5 and Chebyshev’s inequality (see, for example, [15], Appendix, § 7, Theorem 1). AcknowledgmentsThe authors are grateful to the referee for reading the manuscript and making remarks. We are also grateful to Academician S. V. Konyagin for suggesting the idea of the proof of Lemma 15 on simultaneous approximation.
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\Bibitem{KorPop22} |
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