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Q: Is it possible to calculate the integral $$ \int\limits_0^\infty \int\limits_0^\infty\frac{\cos\frac{\pi}2 \left(nx^2-\frac{y^2}n\right)\cos \pi xy}{\cosh \pi x\cosh \pi y}dxdy,~n\in\mathbb{N}\tag{1} $$ using residue theory?

It was proved using Fourier transforms here https://arxiv.org/abs/1712.10324

For example, when $n=3$ $$ \int\limits_0^\infty \int\limits_0^\infty\frac{\cos\frac{\pi}{2} \left(3x^2-\frac{y^2}{3}\right)\cos \pi xy}{\cosh \pi x\cosh \pi y}dxdy=\frac{\sqrt{3}-1}{2\sqrt{6}}. $$ There is a closed form formula to calculate (1) for arbitrary natural $n$, but I don't know how to do it by residue theory. Maybe it is possible in principle, but is residue theory practical in this particular case? It seems such an approach would lead to a sum with $O(n^2)$ terms. Any hints would be appreciated.

Cave Johnson
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  • can you give a reference for the amazing formula $n=3$? i bet it is due to ramanujan – tired Jan 19 '17 at 10:43
  • @tired it is not due to Ramanujan. – Cave Johnson Jan 19 '17 at 10:46
  • thx, that is cool stuff...are do you doing this on a recreational basis or is this for professional purposes? – tired Jan 19 '17 at 11:15
  • @ Nemo: would you mind if I can offer a bounty for this question? – Nicco Oct 11 '17 at 01:12
  • @ Nemo:Based on residue theory.Unfortunately my browser can't display the formulas in your blog – Nicco Oct 11 '17 at 08:08
  • @Nicco I don't mind – Cave Johnson Oct 11 '17 at 08:20
  • Has the case $n=3$ been done with Residue theory? I spent some time thinking about this evening and I don't see how it can be possible. There is no residue theory to my knowledge for double integrals, so you must do a residue calculation on one or both of the iterated integrals, but after some algebra, I see no way to make the LM estimates work, so I don't see how it can be possible.

    If the LM estimates could be worked out, I would be willing to try again.

    – A. Thomas Yerger Jan 08 '19 at 07:26
  • In particular, the place I got stuck is at estimating $| \int_\gamma \frac{\cos \frac{\pi}{2} u^2 \cos \pi u v}{\operatorname{cosh} \pi u/\sqrt{n}}|$ which I obtained by making the change of variables $u = \sqrt{n}x$ and $v = y/\sqrt{n}$, using the angle subtraction formula, distributing, and then pulling out some terms depending only on $v$ from the first integral. This integral is of an even function, so we can extend to an integral over the whole real line. If the LM estimates for a semi-circular arc in the upper half plane could be made to $\to 0$, we could try again to calculate residues. – A. Thomas Yerger Jan 08 '19 at 07:29
  • It is also possible the LM estimates one typically sees are too coarse for this integral, and something better is needed to show that the integral along such an arc goes to $0$ as $R \to \infty$, but I don't have any ideas right now. – A. Thomas Yerger Jan 08 '19 at 07:34
  • @AlfredYerger please see my paper at arxiv regarding 2D Mordell integrals. – Cave Johnson Jan 08 '19 at 16:26
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    This is very interesting. So you would be content if the integrals $I_1$ and $I_2$ just below (26) in this paper would be evaluated with residue theory? https://arxiv.org/pdf/1712.10324.pdf – A. Thomas Yerger Jan 08 '19 at 16:47
  • @AlfredYerger I know $I_{1,2}$ can be evaluated with residue theory. It has been done by Ron Gordon, for exampe, here on MSE. I believe his method can be applied to the case $n=3$ by calculating the integral over $x$ first, then similarly the remaining integral over $y$. – Cave Johnson Jan 09 '19 at 08:13
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    You mentioned that there is a closed form formula. Would you care to provide it? – FShrike Jun 18 '22 at 21:03

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