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What contour and integrand do we use to evaluate $$ \int_0^\infty \frac{x^3}{e^x-1} dx $$

Or is this going to need some other technique?

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up vote 13 down vote accepted

I assume by "or is this going to need some something else" you meant that you are open to non-contour integration techinques, I will show you one of those and let the contour integration be left to someone else (it's not too bad, I just wish I had graphing software).

The classic way to evaluate this integral is as follows

$$\begin{aligned}\int_0^\infty\frac{x^m}{e^x-1}\;dx &= \int_0^{\infty}\frac{e^{-x}x^m}{1-e^{-x}}\;dx\\ &=\int_0^{\infty}\sum_{n=0}^{\infty}x^me^{-x(n+1)}\;\\ &= \sum_{n=0}^{\infty}\int_0^{\infty}x^me^{-x(n+1)}\;dx\\ &=\Gamma(m+1)\sum_{n=0}^{\infty}\frac{1}{(n+1)^{m+1}}\\ &=\Gamma(m+1)\zeta(m+1)\end{aligned}$$

This is, in fact, the way one defines $\zeta(x)$ for $x>1$.

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To solve the integral I thought of a rectangle of vertices $0$, $R$, $R+i2\pi $ and $i2\pi$. Since the function we are going to use is singular in $i2k\pi $ we have to indent the first and the last vertex by using a quarter of circle for each one: let's make them small, of radius $\epsilon$.

enter image description here

Calling this contour $\Gamma$ we have:

$$ \oint_{\Gamma} \frac{z^4}{e^z -1}\mathrm{d}x = 0 $$ by Cauchy's theorem.

As you will notice we have by now 4 segments and 2 quarters of circle to integrate on, namely: $$ \int_\epsilon ^R \frac{x^4}{e^x - 1}\mathrm{d}x + \int_0 ^{2\pi} \frac{(R+iy)^4}{e^{R+iy}- 1}i\mathrm{d}y + \int_{R}^\epsilon \frac{(x+i2\pi)^4}{e^{x+i2\pi}-1}\mathrm{d}x + \int_0 ^{-\frac{\pi}{2}}\frac{(2 \pi i + \epsilon e^{i\theta})^4}{e^{2\pi i + \epsilon e^{i\theta}}-1} i\epsilon e^{i\theta}\mathrm{d}\theta + \int_{2\pi- \epsilon}^{\epsilon}\frac{(iy)^4}{e^{iy}-1}i\mathrm{d}y+ \int_{\frac{\pi}{2}}^{0}\frac{(\epsilon e^{i\theta})^4}{e^{\epsilon e^{i\theta}}-1}i\epsilon e^{i\theta} \mathrm{d}\theta = 0. $$

We can first expand the power in the third integral and note that its first term cancels out with the first integral.

Before writing anything else we take into account the limits $R \to +\infty$, which cancels out the second integral, and $\epsilon \to 0^+$, which annuls the last one and yields a finite value $-8i\pi^5$ in the 4th one.

Therefore we get $$ -i8\pi \int_0 ^\infty \frac{x^3}{e^x - 1} \mathrm{d}x + 24\pi^2\int_0 ^\infty \frac{x^2}{e^x -1}\mathrm{d}x + i 32 \pi^3 \int_0 ^\infty \frac{x}{e^x - 1}\mathrm{d}x- 16\pi^4\int_0 ^\infty \frac{1}{e^x - 1}\mathrm{d}x -i8\pi^5+\frac{i}{2} \int_0 ^{2\pi} y^4 \mathrm{d}y - \frac{1}{2} \int_0 ^{2\pi} \frac{y^4 \sin y}{1-\cos y}\mathrm{d}y=0. $$ Where we have split the 5th integral into its real and imaginary parts.

By considering the expression of the imaginary parts only: $$ -8\pi \int_0 ^\infty \frac{x^3}{e^x - 1} \mathrm{d}x + 32\pi^3 \int_0 ^\infty \frac{x}{e^x - 1}\mathrm{d}x - 8\pi^5 + \frac{16}{5}\pi^5 = 0. $$

The second integral gives $\frac{\pi^2}{6}$ (for a solution of this one a contour similar to the one we've used here is needed; I'm pretty sure it has already been solved here on Math.SE).

Finally: $$ \int_0 ^\infty \frac{x^3}{e^x - 1} \mathrm{d}x = \frac{\pi^4}{8} \left(\frac{16}{3}-8+\frac{16}{5}\right) = \frac{\pi^4}{15}. $$

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Great, thank you – Simon S Oct 30 '14 at 16:42
The second integral after taking the imaginary part equals $\zeta(2)=\pi/6$ and not $\pi/12$, as stated in the answer. The result still does work out though. – Cyclone Apr 15 at 13:13
@Cyclone Thanks, corrected (btw it's $\pi^2 / 6$, not $\pi/6$). – Brightsun Apr 15 at 14:21
Yes, thanks for spotting the typo. – Cyclone Apr 15 at 14:42

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