# Calculating $\int _{0} ^{\infty} \frac{x^{3}}{e^{x}-1}\;dx$

how to calculate $$\int_0^\infty \frac{x^{3}}{e^{x}-1} \; dx$$

Be $q:= e^{z}-1 , p:= z^{3}$ , then $e^{z} = 1$ if $z= 2\pi n i$, so the residue at 0 is : $$\frac{p(z_{0})}{q'(z_{0})} = 2\pi i n ^{3}$$

problem is that this is not symmetric, so how does one find the definite integral?

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Please avoid titles that are entirely in $\LaTeX$. – J. M. Dec 19 '11 at 23:50

In general, \begin{align} \int_0^\infty\frac{x^n}{e^x-1}\mathrm{d}x &=\sum_{k=1}^\infty\int_0^\infty x^ne^{-kx}\mathrm{d}x\\ &=\sum_{k=1}^\infty\frac{1}{k^{n+1}}\int_0^\infty x^ne^{-x}\mathrm{d}x\\ &=\zeta(n+1)\Gamma(n+1) \end{align} In the particular case of $n=3$, we get \begin{align} \int_0^\infty\frac{x^3}{e^x-1}\mathrm{d}x &=\zeta(4)\Gamma(4)\\ &=\frac{\pi^4}{90}\cdot6\\ &=\frac{\pi^4}{15} \end{align}

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Robjohn, thank you. First of all. But is there maybe a way to compute this without the usage of the zeta and the gamma function? Using complex analysis? – VVV Dec 19 '11 at 20:16
Whatever method you use to evaluate this "without $\zeta$" cannot apply also to the nearly identical integral $$B=\int_0^\infty\frac{x^2}{e^x-1}\,dx$$ since that one has value $2\zeta(3)$ . – GEdgar Dec 19 '11 at 22:52
A slight variation, which is small enough to fit in the margin. :) $$\Gamma(s+1)\zeta(s) = \int_0^{\infty} \frac{x^s \, e^x}{(e^x-1)^2} dx.$$ See, for example, this answer and the comment at the end. – Mike Spivey Dec 19 '11 at 23:27
I'll note that this is precisely the integral that comes up when trying to derive the proportionality constant of the Stefan-Boltzmann law for the energy emitted by a blackbody. – J. M. Dec 19 '11 at 23:53
I'm not physicist, you must believe me. Please. – VVV Dec 20 '11 at 4:18

The integral $\int_0^\infty x^s/(e^x-1)\,dx$ with complex $s$ was used by Riemann in one of his proofs of the meromorphic continuation of zeta, using on one hand the expansion written earlier by @robjohn (where the "3" could be replaced by a complex number with positive real part), and, on the other hand, the Hankel or "keyhole" contour, as follows. (Also responding to VVV's comment/question.)

For non-integer complex $s$, at first with positive real part, let $H$ be a Hankel contour: coming in from $+\infty$ along the real line, going clockwise around a small circle around $0$, and back out to $+\infty$. The point is that to have $x^s=e^{s\log x}$ on the integral "out", one has $e^{s(\log x-2\pi i)}$ on the integral "in". Thus, the integral is $\int_H {z^s\,dz\over e^z-1}\cdot {1\over 1-e^{-2\pi is}}$. This gives the meromorphic continuation of $\Gamma(s)\zeta(s)$, for one.

But, also, it evaluates $\zeta(s)$ at non-positive integers: rearranging, $$\zeta(s+1) = {1\over \Gamma(s+1)\cdot (1-e^{-2\pi i s})}\cdot \int_H {z^s\,dz\over e^z-1}$$ The leading term has a removable singularity at negative integers, and the integral can be evaluated by residues.

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I had trouble seeing this until I found the following approach--late and for what it is worth. I like that it shows the summation explicitly (and reminds us of the expressions for Gamma and Zeta).

$\Gamma(x) = \int_{0}^{\infty}t^{x-1}e^{-t}dt$ for x > 0. Make the variable substitution t = ru:

$\Gamma(x) = \int_{0}^{\infty}(ru)^{x-1}e^{-ru}r\ du =r^x \int_{0}^{\infty}u^{x-1}e^{-ru}du$

So that

$\frac{1}{r^x} = \frac{1}{\Gamma(x)}\int_{0}^{\infty}u^{x-1}e^{-ru}du$.

$\zeta(x) = \sum_{1}^{\infty}\frac{1}{r^x}= \frac{1}{\Gamma(x)}\sum_{1}^{\infty} \int_{0}^{\infty}u^{x-1}e^{ru}du$ = $\frac{1}{\Gamma(x)}\int_{0}^{\infty}u^{x-1}\sum_{1}^{\infty}e^{-ru}du$.

Finally,

$\zeta(x) = \frac{1}{\Gamma(x)}\int_{0}^{\infty}u^{x-1}\frac{ e^{-u}}{1-e^{-u}}du$ and so $\Gamma(x)\zeta(x) = \int_{0}^{\infty}\frac{u^{x-1}}{e^u-1}du$

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