# Evaluating the definite integral $\int_0^\infty \frac{\mathrm{e}^x}{\left(\mathrm{e}^x-1\right)^2}\,x^n \,\mathrm{d}x$

I am having difficulty evaluating definite integrals of the form $\int_0^\infty \frac{\mathrm{e}^x}{\left(\mathrm{e}^x-1\right)^2}\,x^n \,\mathrm{d}x$. I would appreciate any guidance that could be offered. I am aware that these evaluate to constant powers of $\pi$, but find the integration challenging. Thank you.

## 1 Answer

$$\int_0^\infty \frac{e^x}{(e^x-1)^2} x^n dx = \int_0^\infty \frac{e^{-x}}{(1-e^{-x} )^2} x^n dx .$$

For $|z|<1$ we have $\displaystyle \frac{1}{1-z} =\sum_{k=0}^{\infty} z^k$ so by differentiating we get $\displaystyle \frac{1}{(1-z)^2} = \sum_{k=1}^{\infty} k z^{k-1}.$ Thus for $x\in (0,\infty),$ $$\frac{e^{-x}}{(1-e^{-x} )^2} = \sum_{k=1}^{\infty} k e^{-kx}.$$ Hence, (after applying the monotone convergence theorem) the desired integral is equal to $$\sum_{k=1}^{\infty} k \int^{\infty}_0 x^n e^{-kx} dx .$$

If we let $u=kx$ we find that $$\int^{\infty}_0 x^n e^{-kx} dx = \int^{\infty}_0 \left( \frac{u}{k} \right)^n e^{-u} \cdot \frac{1}{k} du = \frac{1}{k^{n+1}} \Gamma(n+1) = \frac{n!}{k^{n+1}}.$$

Thus, $$\int^{\infty}_0 \frac{e^x}{(e^x-1)^2} x^n dx = n! \zeta(n) .$$

Your suspicion that the value of the integral is a constant times a power of $\pi$ is correct for even integers, as $$(2n)! \cdot \zeta(2n) = (2n)! \cdot (-1)^{n+1} \frac{B_{2n} (2\pi)^{2n} }{2 (2n)!} = (-1)^{n+1} \frac{B_{2n} (2\pi)^{2n} }{2}$$

but for odd n no simpler form for $\zeta(n)$ is known (and it probably isn't simple powers of $\pi.$)

• Thank you for detailing your approach. I will study it carefully as it contains much new math for me. Does the monotone convergence theorem refer simply to the summation of an infinite geometric series? I notice that many of the questions I have asked about integration have been solvable with recurrence relations. Supposing one might have encountered these in a formal educational setting, in what sort of class might they have been included? Analysis? I was not exposed to such functions in univariate or multivariate calculus classes. – user001 Mar 7 '12 at 15:24
• There are a few 'monotone convergence theorems' so I guess I should have been more specific. You may be thinking about 'bounded monotone sequences converge' but I am talking about the theorem encountered when one studies measure theory (analysis). If you don't know it, don't worry about it, it's use here in my proof is to justify the switching of the summation and integral operators, which I hope you can believe since switching finite sums and integrals is of course fine. – Ragib Zaman Mar 7 '12 at 15:37
• Finding recurrence relations for integrals is often seen when one first studies integration by parts, as that is one of the main tools for doing so. It should have been somewhere in your single variable calculus course. You can try it yourself right now though, I'll give you some exercises: 1) Integrating by parts, find a recurrence for $I_n = \int^{\infty}_0 x^n e^{-x} dx.$ Can you solve this recurrence to find an explicit form for $I_n$? 2) Let $J_n = \int^{\infty}_1 \frac{du}{u^{n+1} (1+u) } .$ Find a recurrence for $J_n$ and hence find $J_4.$ (This recurrence can't be solved explicitly). – Ragib Zaman Mar 7 '12 at 15:41
• 3) Let $K_n = \int^{\pi/2}_0 \sin^n x dx.$ Find a recurrence relation for $K_n$ and hence evaluate $K_5.$ Try to find a closed form solution of the recurrence. – Ragib Zaman Mar 7 '12 at 15:45
• That is very kind of you. I will work on these momentarily. – user001 Mar 7 '12 at 16:01