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When describing black body radiation in Physics, one comes across an integral of the form

$$I(a)=\int_0^\infty\frac{x^3}{e^{ax}-1}dx \tag1$$

for $a>0$. In our lecture, we were given that this integral evaluates to

$$I(a)=\frac{\pi^4}{15a^4}$$

without a derivation of this result.

While reviewing my course notes over the Christmas break, I came across this integral and have tried to evaluate it myself, but I have not been successful so far.

My first step, of course, was to substitute $u:=ax$, which gives us

$$I(a) = \frac{1}{a^4}\int_0^\infty\frac{u^3}{e^u-1}du \tag2$$

but beyond that, I'm stuck.

I've tried differentiation under the integral sign on the original integral in $(1)$, which lead to the integral

$$\int_0^\infty\frac{\log^4 u}{(u-1)^2}du$$

which I also no have no clue how to tackle.

I've also thought about using contour integration, but I'm unsure as to what would be an appropriate contour to consider. I know that the integrand has singularities at $z=2\pi ik$ for $k\in\mathbb{Z},k\neq 0$ but I'm not even sure how one would go about calculating the residues at these singularities due to the peculiar form of the denominator, which is unlike any I have previously encountered.

How would one go about deriving that

$$\int_0^\infty\frac{x^3}{e^{ax}-1}dx = \frac{\pi^4}{15a^4}$$

is true?

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    See [How was the integral for Zeta Function created](http://math.stackexchange.com/questions/1848949/how-was-the-integral-for-zeta-function-created) for the general case (first make the substitution $u = ax$ to get it on that form). See also: [Calculating $ \int _{0} ^{\infty} \frac{x^{3}}{e^{x}-1}\;dx$](http://math.stackexchange.com/questions/92793/calculating-int-0-infty-fracx3ex-1-dx) – 2017-01-01
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    Thanks. Since my question doesn't offer anything new, should I delete it? – 2017-01-01
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    @Tom perfectly up to you. No way you would've known this on your own if you asked this question, and I'm sure plenty of others don't either, so you could leave it/close as duplicate. – 2017-01-01
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    You can't (and should not) delete a question after getting an answer. If you feel one the answers linked are satisfactory it can be marked as a duplicate (then the question stays but is linked to that post - this makes it easier in the future for people to find relevant posts). – 2017-01-01
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    @Winther Listen to the wiser person. – 2017-01-01

1 Answers 1

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Use the substitution $ax=u$.

$$\int_0^\infty\frac{x^3}{e^{ax}-1}\ dx=\frac1{a^4}\int_0^\infty\frac{x^3}{e^x-1}\ dx$$

Now, recalling an integral form of the zeta function, we then have

$$\frac1{a^4}\int_0^\infty\frac{x^3}{e^x-1}\ dx=\frac{\zeta(4)3!}{a^4}=\frac{\pi^4}{15a^4}$$

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    @Tom No problem, cheers! :-D – 2017-01-01