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Copy cat of Ramanujan identity

It has a neat closed form

$$\displaystyle\int_{-\infty}^{+\infty}xe^{-{\pi\over 4}x^2}{\sinh^2(3\pi x)\over \sinh(\pi x)}\mathrm dx=4\left(e^{\pi}+3e^{9\pi}+5e^{25\pi}\right)\tag1$$

Making an attempt:

Recall $$\int\sinh^2(3\pi x)\mathrm dx={\sinh(3\pi x)\cosh(3\pi x)\over 6\pi}-{x\over 2}+C\tag2$$

$$\int{1\over \sinh(\pi x)}\mathrm dx={1\over \pi}\ln{\tanh{\pi x\over 2}}+C\tag3$$

$$\int xe^{-{\pi\over 4}x^2}\mathrm dx=-{2\over \pi}e^{-{\pi\over 4}x^2}+C\tag4$$

$$\sinh(3\pi x)=3\sinh(\pi x)+4\sinh^3(\pi x)\tag5$$

Applying $(5)$ then $(1)$ becomes

$$\int_{-\infty}^{+\infty}xe^{-{\pi\over 4}x^2}{\sinh(3\pi x)}[3+4\sinh^2(\pi x)]\mathrm dx\tag6$$

Rewrite $(5)$ as

$$3\int_{-\infty}^{+\infty}xe^{-{\pi\over 4}x^2}{\sinh(3\pi x)}\mathrm dx+4\int_{-\infty}^{+\infty}xe^{-{\pi\over 4}x^2}{\sinh^2(\pi x)}\mathrm dx=I_1+I_2\tag7$$

Applying IBP to $I_1$, then we have

$$I_1=-{2\over \pi}e^{-{\pi\over 4}x^2}{\sinh(3\pi x)}+\int e^{-{\pi\over 4}x^2}{\cosh(3\pi x)}\mathrm dx\tag8$$

and to $I_2$

$$I_2=-{2\over \pi}e^{-{\pi\over 4}x^2}{\sinh^2(\pi x)}+{2\over \pi^2}\int e^{-{\pi\over 4}x^2}{\sinh(2\pi x)}\mathrm dx\tag9$$

This is too lengthy of a calculation method.

How can we tackle $(1)$ in a less cumbersome way? Or else just prove $(1)$

Community
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gymbvghjkgkjkhgfkl
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1 Answers1

6

Hint. One may observe that $$ {\sinh^2(3\pi x)\over \sinh(\pi x)}=\frac12e^{5\pi x}-\frac12e^{-5\pi x}+\frac12e^{3\pi x}-\frac12e^{-3\pi x}+\frac12e^{\pi x}-\frac12e^{\pi x} $$ giving $$ \begin{align} &\int_{-\infty}^{+\infty}xe^{-{\pi\over 4}x^2}{\sinh^2(3\pi x)\over \sinh(\pi x)}\mathrm dx \\\\=&\int_{-\infty}^{+\infty}xe^{-{\pi\over 4}x^2}\left(\frac12e^{5\pi x}-\frac12e^{-5\pi x}+\frac12e^{3\pi x}-\frac12e^{-3\pi x}+\frac12e^{\pi x}-\frac12e^{\pi x}\right)\mathrm dx \end{align} $$ then one may conclude with the gaussian result $$ \int_{-\infty }^{\infty } x e^{-\frac{\pi x^2}{4}+n x} \, dx=\frac{4}{\pi }\cdot n\cdot e^{\frac{n^2}{\pi }}, \quad n=0,\pm1,\pm2,\cdots $$ obtained by differentiating $$ \int_{-\infty }^{\infty }e^{-\frac{\pi x^2}{4}+a x} \, dx=2 e^{\large \frac{a^2}{\pi}}, \quad a \in \mathbb{R}, $$ with respect to $a$.

Olivier Oloa
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