$$J = \int\limits_0^\infty\frac{x^3}{e^x-1}\mathrm dx$$

Show that $J = \pi^4/15$.

Question

$$J = \int\limits_0^\infty\frac{x^3}{e^x-1}\mathrm dx$$

Show that  $J = \pi^4/15$.

zarkon · Answer

you can use

$$\frac{1}{e^{x}-1}=\frac{e^{-x}}{1-e^{-x}}=\sum_{k=1}^{\infty}e^{-kx}$$
...
convert the problem into finding $\zeta(4)$

unklerhaukus · Answer

... where $\zeta$ is the Riemann zeta function

kainui · Answer

I was thinking I could find the poles and turn this into a contour integral, but I haven't had enough practice with them to know how to do this. There's the clear way in which $e^{ix}-1=0$ has solutions for when $x=2 \pi n$ except at the origin it looks like it isn't a pole since by L'Hopital's rule it's 0 there.

I then tried to look at when $ \frac{e^z-1}{z^3} =  0$ to find the poles to see if there was some kinda simple way to get the residue(s) by turning it into a Laurent series cause I really don't know what I'm doing lol.

$$\frac{e^z-1}{z^3} = \sum_{n=1}^\infty \frac{z^{n-3}}{n!} = \sum_{k=-2}^\infty \frac{z^k}{(k+3)!}$$

Unfortunately this is not the Laurent series of our function so the k=-1 term is not the residue and I'm   basically just sorta fooling around. I could go back to my original thing and do a kind of contour integral maybe like this? |dw:1458265937682:dw|

Those are the $2 \pi i n$ poles and maybe let this thing go out to infinity and hope the arc part goes to 0 as it goes to infinity since there's an exponential in the denominator I believe this much is true. I just don't know about this return path and the fact that I'm including infinitely many poles.

Oh well I'll keep thinking about this. Maybe I can just do a single contour containing one pole or alternate between poles?

owen3 · Answer

I just wanted to mention that
$$\zeta (4) = {\frac {{\pi }^{4}}{90}}$$

amilapsn · Answer

Can we do this with double or some higher dimensional integral? Like finding $\int_0^\infty \frac{\sin x}{x}dx$?
If we can, is there a method to think it?