I was integrating the Planck distribution and came across this integral:

$$\int_0^\infty \frac {1}{x^5\cdot \left(e^{\frac{a}{x}}-1\right)} dx$$

From this page: http://w.astro.berkeley.edu/~echiang/rad/ps1ans.pdf the given solution is

$$\int_0^\infty \frac {1}{x^5\cdot \left(e^{\frac{1}{x}}-1\right)}dx = \frac {\pi^4}{15} $$

and I could analytically work out that the general solution is

$$\int_0^\infty \frac {1}{x^5\cdot \left(e^{\frac{a}{x}}-1\right)}dx = \frac {\pi^4}{15a^4} $$

I was wondering how one could prove this, and if my general solution is accurate or not.

  • 4
    $\begingroup$ Did you not just compute it? Why do you think your computation might be wrong? $\endgroup$ – Paul Oct 1 '19 at 13:21
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    $\begingroup$ Missing $dx$ in equations. $\endgroup$ – StephenG Oct 1 '19 at 13:28

We can do something more generally for $a>0, b>2$: $$\int_0^\infty \frac {1}{x^{b}\left(e^{\frac{a}{x}}-1\right)}dx\overset{\frac{a}{x}=t}=\frac{1}{a^{b-1}}\int_0^\infty \frac{t^{b-2}}{e^t-1}dt=\frac{\zeta(b-1)\Gamma(b-1)}{a^{b-1}}$$ Above follows using the integral definition for the zeta function.

And indeed for the case $b=5$ we obtain the announced result.

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