Estimate of integral of Fourier series

Question

Not really my area, so I appeal to you: any idea on how to estimate $$\zeta^{-k}(1+\alpha)\int_0^1 \left(\sum_{q=1}^\infty \cos(2\pi q x)q^{-1-\alpha}\right)^k dx$$ for $k\in\mathbb{N}$ and $0<\alpha<1$. I mean anything better than 1. I tried various things: expanding the powers and using uniformity of the convergence of the series to exchange with integral sign and see if enough cosine integrals get zero, but the approach failed for too many repeated terms. The series does not seems to be coming from an elementary function either. My aim is to run a sum of these terms over $k$ and get convergence. Technicalities aside my gut feeling is that the cosine should reduce the series significantly enough when compared to zeta, but also proving uniform boundedness away from one failed. Could something related to Riemann-Lebesgue be lurking here? Still checking. I’m open to any suggestion... cheers!

Answer 1

As far as I see, you're interested in the asymptotics of the integral (for $\newcommand{\lylog}{\operatorname{Li}}k\to\infty$). We rewrite it as $$\int_0^1\big(f_\alpha(x)\big)^k dx=2\int_0^{1/2}\big(f_\alpha(x)\big)^k dx$$ and observe that the main contribution comes from the neighborhood of $x=0$.

In terms of the polylogarithm, $$2\zeta(1+\alpha)f_\alpha(x)=\lylog_{1+\alpha}(e^{2i\pi x})+\lylog_{1+\alpha}(e^{-2i\pi x}),$$ and now we use the expansion of $\lylog_s(e^z)$ for $|z|<2\pi$ cut along the positive real axis: $$\lylog_{1+\alpha}(e^z)=\Gamma(-\alpha)(-z)^\alpha+\sum_{n=0}^{\infty}\zeta(1+\alpha-n)\frac{z^n}{n!},$$ which, after substitution and application of the functional equation for $\zeta$, yields $$f_\alpha(x)=1+\frac{x^\alpha}{2\zeta(-\alpha)}+\sum_{n=1}^{\infty}(-1)^n\frac{\zeta(1+\alpha-2n)}{\zeta(1+\alpha)}\frac{(2\pi x)^{2n}}{(2n)!}\label{mainexp}\tag{*}$$ The first two terms give the main asymptotics. Informally, we substitute $x=\big(-2\zeta(-\alpha)t/k\big)^{1/\alpha}$: $$\int_0^{1/2}\big(f_\alpha(x)\big)^k\,dx=\frac1\alpha\left(-\frac{2\zeta(-\alpha)}{k}\right)^{1/\alpha}\int_0^{t_k}t^{1/\alpha-1}\left(1-\frac{t}{k}+\mathcal{O}\left(\left(\frac{t}{k}\right)^{2/\alpha}\right)\right)^k dt,$$ where $t_k=-k/\big(2^{1+\alpha}\zeta(-\alpha)\big)\to+\infty$ as $k\to\infty$; by DCT, the integral converges to $$\int_0^\infty t^{1/\alpha-1}e^{-t}\,dt=\Gamma(1/\alpha).$$

Finally $$\int_0^1\big(f_\alpha(x)\big)^k dx\underset{k\to\infty}{\asymp}\color{blue}{C_\alpha k^{-1/\alpha}},\qquad C_\alpha=2\big(-2\zeta(-\alpha)\big)^{1/\alpha}\Gamma(1+1/\alpha).$$ More detailed asymptotics can be [tediously] obtained from the remaining terms of $\eqref{mainexp}$.