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

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.

• Try integrating by parts, taking derivatives with respect to sine and integrating with respect to sinh Aug 4, 2020 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 $$2t \int_{-\infty}^{\infty}x \coth{(\pi x)} \cos{(t \pi x^2)} \; \mathrm{d}x$$
– user801111
Aug 4, 2020 at 21:31
• $a/b$ has to be rational numbers. This is connected to Gauss sums (Ramanujan). Contour integral won't work for more complex cases. Aug 5, 2020 at 4:21
• @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. Aug 5, 2020 at 6:57
• @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. Aug 5, 2020 at 7:33