I am trying to evaluate the integral $$ \int_{0}^{\infty} \frac{x \exp\left\{-\beta^2 x^2\right\}}{\sinh \left(\frac{\pi x}{2}\right)} \mathrm{d} x, $$ where $\beta \in \mathbb{R}$ is some constant.

After many unfruitful attempts and checking up several integral handbooks, I am still not sure how to handle it. Does there exist an analytic solution?

Thanks in advance!


Inspired by the discussions in this question, I tried to make the following manipulation (assume we can interchange the order of the infinite sum and the definite integral): $$ \begin{aligned} \int_{0}^{\infty} \frac{x \exp(-\beta^2 x^2)}{\sinh(\pi x/2)}\mathrm{d}x =& \frac{16}{\pi^3} \int_{0}^{\infty}\sum_{n=0}^{\infty} x \exp(-\beta^2x^2) \exp(-(2n+1)x) \mathrm{d} x \\ =&\frac{16}{\pi^3} \sum_{n=0}^{\infty} \exp\left(\frac{(2n+1)^2}{4\beta^2} \right) \int_{0}^{\infty} x \exp\left\{ -\frac{1}{2\frac{1}{2\beta^2}} \left(x + \frac{2n+1}{2\beta^2} \right)^2 \right\} \mathrm{d}x\\ =& \frac{16}{\pi^3}\sum_{n=0}^{\infty} \exp\left(\frac{(2n+1)^2}{4\beta^2}\right) \left\{ \frac{\sqrt{2\pi}}{2\beta^2}\exp\left(-(2n+1)^2\beta^2\right) \right. \\ &\left.- \frac{\sqrt{\pi}(2n+1)}{2\beta^3} \left(1-\Phi\left(\frac{\sqrt{2}}{2}(2n+1)\beta\right)\right) \right\} \end{aligned}, $$ where $\Phi(\cdot)$ is the standard normal cdf. This still seems not very promising: I have little idea how to deal with this infinite series.

  • $\begingroup$ I tried expanding the integrand in a power series in $\beta$, but the resulting power series does not converge. $\endgroup$ May 31, 2019 at 14:40

1 Answer 1


If we name the OP integral $J$ then let me introduce another integral:

$$I= \frac{\pi}{2} \int_{-\infty}^\infty \frac{x e^{- \beta^2 x^2}}{\sinh (\frac{\pi}{2} x)} dx=\pi J$$

We can use the following result:

$$\int_0^\infty \frac{\cos (x t)}{\cosh^2 t}dt=\frac{\frac{\pi}{2} x}{\sinh (\frac{\pi}{2} x)}$$

Which leads to:

$$I=\int_0^\infty \int_{-\infty}^\infty \frac{\cos (x t) e^{- \beta^2 x^2}}{\cosh^2 t} dx dt=2 \sqrt{\pi} \int_0^\infty \frac{e^{- t^2}}{\cosh^2 (2 \beta t)} dt$$

Where we used a well known result:

$$\int_{-\infty}^\infty \cos (a x) e^{- x^2} dx=\sqrt{\pi} e^{-a^2/4}$$

Using integration by parts for the last integral, we get a different form:

$$I= \frac{2 \sqrt{\pi}}{\beta} \int_0^\infty t e^{-t^2} \tanh(2 \beta t) dt$$

Using geometric series:

$$\tanh(2 \beta t)=\left(2 \sum_{n=0}^\infty (-1)^n e^{-4 \beta n t}\right)-1$$

Substituting and taking the integrals (I omit this part), we obtain the following series:

$$I=\frac{2 \sqrt{\pi}}{\beta} \sum_{n=0}^\infty (-1)^n \left(1-2 \sqrt{\pi} \beta n ~e^{4 \beta^2 n^2} \operatorname{erfc} (2 \beta n) \right)-\frac{\sqrt{\pi}}{\beta} $$

Where $\operatorname{erfc}=1-\operatorname{erf}$ is the complementary error function.

The series shows good convergence, even though it looks awkward.

Note that the OP already considered a similar form of the integral in another question: Series Expansion of $\int_{\mathbb{R}} \tanh\left(\beta x\right) x \exp\left(-\frac{x^2}{2}\right) \mathrm{d} x$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.