Asymptotic formula for the moments of Bernoulli convolutions

For each $\lambda$, $0<\lambda<1$, we define a random variable
$$ Y_\lambda = (1-\lambda)\sum_{n=0}^\infty \xi_n\lambda^n, $$
where $\xi_n$ are independent random variables with
$$ \mathrm{P}\{\xi_n =0\} =\mathrm{P}\{\xi_n =1\} =\frac12. $$
The distribution of $Y_\lambda$ is called a symmetric Bernoulli convolution. The main result of this paper is
$$ M_n = \mathrm{E} Y_\lambda^n = n^{\log_{\lambda}2} 2^{\log_\lambda(1-\lambda)+0.5\log_\lambda2-0.5} e^{\tau(-\log_{\lambda}n)}\left(1 + \mathcal{O}(n^{-0.99})\right), $$
where
$$ \tau(x)=\sum_{k\ne0}\frac1k\alpha\left(-\frac{k}{\ln\lambda}\right)e^{2\pi ikx} $$
is a 1-periodic function,
$$ \alpha(t) = -\frac{1}{2i\mathrm{sh}\,(\pi^2t)} (1-\lambda)^{2\pi i t}(1 - 2^{2\pi i t})\pi^{-2\pi i t }2^{-2\pi i t }\zeta(2\pi i t), $$
and $\zeta(z)$ is the Riemann zeta function.
The article is published in the author's wording.