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Evaluate $$\int_{-\infty}^{\infty} \frac{\sin{\left(t\pi x^2\right)}}{\sinh^2{\left(\pi x\right)}} \; \mathrm{d}x$$

I converted $\sinh{(x)}$ to exponential form and considered Imaginary part of the numerator: $$4\Im{\left(\int_{-\infty}^{\infty} \frac{e^{2 \pi x}e^{t\pi x^2}}{{\left(e^{2 \pi x} -1\right)}^2} \; \mathrm{d}x\right)}$$ I think a semi circle contour in upper quadrants would work. Residues are at $x=k \cdot i, k \in \mathbb{N}$ including $0i$. Where I calculated the residues to be $$-\frac{2t}{\pi} \sum_{n=0}^{\infty} ne^{-n^2 \pi i t}$$ I dont know what to do from here (closed form) or maybe my work is wrong? Ideas or tips please.

Quanto
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    Try integrating by parts, taking derivatives with respect to sine and integrating with respect to sinh – Moko19 Aug 04 '20 at 21:21
  • @Moko19 I just tried it and I'm stuck on $$\frac{1}{2 \pi}\lim_{x \to -\infty} \frac{\sin{(t \pi x^2)}}{e^{2 \pi x}-1}+ t \int_{-\infty}^{\infty} \frac{x \cos{(t \pi x^2)}}{e^{2 \pi x}-1} ; \mathrm{d}x$$ I tried contour integration again with the new integral but get the same sum. Is this what you meant?

    Also integral is $$2t \int_{-\infty}^{\infty}x \coth{(\pi x)} \cos{(t \pi x^2)} ; \mathrm{d}x$$

     –  Aug 04 '20 at 21:31
  • https://math.stackexchange.com/questions/61605/tough-integral-involving-sinx2-and-sinh2-x?noredirect=1&lq=1 – Quanto Aug 04 '20 at 21:45
  • @Quanto How would I adjust $f(z)$? I know this is wrong, but if I use $$f(z)=\frac{\cos\left(t \pi z^2\right)}{\sinh(2\pi z)\sinh^2(\pi z)}$$ or is this wrong? Using this $f(z)$ I got $$I=\left(\frac{1}{2} + \cos{\left(\frac{\pi t}{4}\right)}+ \frac{\pi(2t^2-1)\cos{(\pi t)}+t\sin{(\pi t)}}{2\pi}\right)$$ where $t=\frac{b}{a}$ but this is wrong according to values of $t$ in Wolfram (only $t=1$ is right). –  Aug 04 '20 at 23:17
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    $a/b$ has to be rational numbers. This is connected to Gauss sums (Ramanujan). Contour integral won't work for more complex cases. – Infiniticism Aug 05 '20 at 04:21
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    @User628759 Indeed. The integral is an easy generalization of techniques here. The answer involves Gauss sum, which in some cases, can be simplified non-trivially. – pisco Aug 05 '20 at 06:57
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    @pisco I remenber Ramanujan evaluates a lot of them, with values of form $P(\frac1 \pi)$, $P$ polynomial. I generalized some of them to arbitrary weights a year ago, so I decided not to post this again. : ) Your technique is also ingenious. – Infiniticism Aug 05 '20 at 07:33

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