I met this integral in my group:

$$\int_0^\infty \frac{x}{(e^x-e^{-x})^m}dx\qquad 0<m<2$$

I tried: $$\int_0^\infty \frac{x}{(e^x-e^{-x})^m}dx=\int_0^\infty \frac{xe^{-mx}}{(1-e^{-2x})^m}dx$$

let $e^{-2x}=t$,$x=\frac{-\ln t}{2}$ so the integral is equal to:

$$\int_1^0\frac{(\frac{-\ln t}{2})t^{\frac m2}}{(1-t)^m}(-\frac{1}{2t})dt=-\frac14\int_0^1(1-t)^{-m}t^{\frac m2-1}\ln tdt $$

But how to calculate this new integral? Who can help me. Thanks !

  • $\begingroup$ So $m$ is not necessarily an integer? $\endgroup$ – robjohn Oct 19 '15 at 7:31

Assume that $0<m<1$.

Let $$ F(s)=\int_0^{+\infty}\frac{1}{(1-e^{-2x})^m}e^{-s x}\,dx. $$ Then, your integral equals $-F'(m)$.

The substitution you already did shows that $$ F(s)=\frac{1}{2}\int_0^1 (1-t)^{(1-m)-1}t^{s/2-1}\,dt=\frac{1}{2}\beta(1-m,s/2). $$ What remains is to calculate $$ -F'(m)=-\frac{d}{ds}\frac{1}{2}\beta(1-m,s/2)\Bigr|_{s=m} $$ But (see here for example) $$ \frac{d}{dx}\beta(y,x)=\beta(y,x)\bigl[\psi_0(x)-\psi_0(x+y)\bigr], $$ where $\psi_0$ denotes the Polygamma function. Hence, $$ -F'(m)=-\frac{1}{4}\beta(1-m,m/2)\bigl[\psi_0(m/2)-\psi_0(1-m/2)\bigr] $$ By the reflection principle for $\psi_0$ $$ \psi_0(1-z)-\psi_0(z)=\pi\cot(\pi z), $$ we find that

$$ \int_0^{+\infty}\frac{x}{(e^x-e^{-x})^m}\,dx=\frac{\pi}{4}\beta(1-m,m/2)\cot(m\pi/2). $$


As kindly pointed out to me by @JanG, the original integral $F(s)$ diverges for $1\leq m<2$ (I was originally not reading the question well, and just assumed $0<m<1$).

To extend the equality in the yellow box above to $1\leq m<2$, we might argue as follows: The integral in the left-hand side is convergent for $0<\text{Re}\,m<2$, and depends analytically on $m$. The right-hand side depends analytically on $m$ (note that the singularity at $m=1$ of the beta function is canceled by the cotangent), and can be analytically continued from $0<\text{Re}\,m<1$ to $0<\text{Re}\,m<2$. Since the expressions agree for $0<m<1$, they agree for all $0<\text{Re}\,m<2$, by the identity theorem of analytic functions.

  • $\begingroup$ Nice!~~~Thanks very much! $\endgroup$ – smallsmallice Oct 19 '15 at 9:48

Hint: $$\int_0^\infty \frac{x}{(e^x-e^{-x})^m}dx=2^{-m}\int_0^\infty x\sinh^{-m}xdx= \quad2^{-m}\int_0^\infty \mathrm{arcsinh}(y) y^{-m} \frac{1}{\sqrt{1+y^2}}dy $$

I guess you have to look at I. S. Gradshteyn and I. M. Ryzhik. Table of Integrals, Series, and Products. Edited by A. Jeffrey and D. Zwillinger. Academic Press, New York, 7th edition, 2007. You may either find there the solution you need or at least to find an idea of what to do...


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