Asymptotics of $I_n:=\int_1^\infty \frac{d}{dx} [x^{3/2}\psi ' (x)]x^{-1/4}(\log x)^{2n}$ as $n\to\infty$.

Question

In Section 1.8(3) of the monograph Riemann zeta function, author H. M. Edwards says that Riemann derived the power series $$\xi(s)=\sum_{n=0}^\infty a_{2n}\left(s-\frac{1}{2}\right)^{2n}$$ and stated that this series converges very rapidly, but didn't give any explicit estimates. Here, $\xi$ is the Riemann xi function and the coefficients $a_{2n}$ are given by \begin{align*} a_{2n} &= 4\int_1^\infty \frac{d}{dx}[x^{3/2}\psi'(x)]x^{-1/4}\frac{\left(\frac{1}{2}\log x\right)^{2n}}{(2n)!}dx\\ &=\frac{4}{2^{2n}(2n)!}\int_1^\infty \frac{d}{dx}[x^{3/2}\psi'(x)]x^{-1/4}\left(\log x\right)^{2n}dx \end{align*} where $\psi$ is given by the series $$\psi(x)=\sum_{n=1}^\infty e^{-n^2\pi x}$$ I'm interested in how quickly the series for $\xi$ converges, so I need to determine the asymptotics of the integrals $$I_n=\int_1^\infty \frac{d}{dx}[x^{3/2}\psi'(x)]x^{-1/4}\left(\log x\right)^{2n}dx$$ To check for any exponential or hyper-exponential growth, I computed the first $100$ terms of $\frac{I_{n+1}}{I_n}$ with Mathematica. It seems like these quotients grow without bound as $n\to\infty$ but the rate of increase decreases, with the numbers suggesting that $\frac{I_{n+1}}{I_n}=o(n)$. If both of these are true, then the $I_n$'s grow quicker than any exponential but $I_n=O(\varepsilon^n n!)$ for any $\varepsilon>0$.

If these conjectures are correct (are they?), they give me a good estimate of the rate of decay of the $a_{2n}$'s, but I'd like to know this rate precisely. Thus, can we determine the asymptotics of the $I_n$'s? Any help is appreciated.

Answer 1

In this paper, the author presents the following asymptotic growth rate for the coefficients: $$ (2n)!\,a_{2n} = \mathcal{O}(1)\,n^{9/4}\left(\frac{\log n}{2}\right)^{2n - 3/4} \exp \left( - \frac{2n}{\log n} \right), \qquad n\to+\infty. $$ Using Stirling's formula, $$ a_{2n} = \mathcal{O}(1)\,n\left( {\frac{{\log n}}{{4n}}} \right)^{2n - 3/4} \exp \left( {2n\left( {1 - \frac{1}{{\log n}}} \right)} \right). $$

Answer 2

Too long for a comment

Using the heuristic approach, we can find the leading asymptotic term at $n\to\infty$ of the integral $$I(a,l,n)=\int_1^\infty x^l e^{-ax} \ln^n(x)dx$$ The solution is nearly the copy-past of the solution here. Using the same logic, we get $$I(a,l,n)\sim\sqrt\frac{2\pi n}{1+W\big(\frac na\big)}\,\frac{e^{l\,W(\frac na)}}a\left(W\big(\frac na\big)\right)^ne^{-ae^{W(\frac na)}}$$ where $W$ denotes Lambert W function. The approximation works rather well (please, see below), providing the constant ($O(1)$ in the @Gary' post) as well. For the rough estimation we can take $W(x)=\ln x-\ln\ln x+o(1); x\to\infty$, but if we consider $W(x)$, the accuracy is higher.

We see that the function declines exponentially if $a$ is growing; it means that we can consider only first term of $\psi(x)=\sum_{n=1}^\infty e^{-n^2\pi x}\approx e^{-\pi x}$. We also see that the function is growing exponentially with $l$. Hence, we just can take $$\frac{d}{dx}[x^{3/2}\psi'(x)]\approx -\pi^2x^{3/2}e^{-\pi x}$$ Taking $a=\pi; l=\frac54$, and switching $n\to 2n$, in your notations, $$\frac1{\pi^2}I_n=\frac1{\pi^2}\int_1^\infty \frac{d}{dx}[x^{3/2}\psi'(x)]x^{-1/4}\left(\log x\right)^{2n}dx\sim I\big(\pi;\frac54;2n\big)$$ $$\sim \sqrt\frac{4\pi n}{1+W\big(\frac {2n}\pi\big)}\,\frac{e^{\frac54W(\frac {2n}\pi)}}\pi\left(W\big(\frac {2n}\pi\big)\right)^{2n}e^{-\pi e^{W(\frac {2n}\pi)}}$$ As I mentioned, there is a decent numeric agreement (WA check). For example, $\displaystyle\quad l=\frac54; a=\pi; n=686.10...\,\Rightarrow \,W(\frac {n}\pi)=4$ $$I(a, l, n)=5.3279...\times10^{341};\quad approximation=5.2851...\times10^{341}$$ $\displaystyle\quad l=\frac54; a=\pi; n=2331.26...\,\Rightarrow \,W(\frac {n}\pi)=5$ $$I(a, l, n)=7.9746...\times10^{1430};\quad approximation=7.9507...\times10^{1430}$$ etc.