Prove that $\int_0^{\infty} \frac{x^{2n+1}}{e^{\pi x}-1}\mathrm{d}x \in \mathbb{Q}, \forall \,n\in\mathbb{N}$ Prove that
$$\int_0^{\infty} \frac{x^{2n+1}}{e^{\pi x}-1}\mathrm{d}x \in \mathbb{Q}, \forall \,n\in\mathbb{N}$$
 A: This is an supplement of rlgordonma's answer if one doesn't want to use Bernoulli's numbers.
Start with Euler's infinite product expansion of $\sin(x)$:
$$\sin(x) = x \prod_{k=1}^{\infty} \left( 1 - (\frac{x}{k\pi})^2\right)$$
We have for $|x| < \pi$,
$$
\log \frac{\sin(x)}{x} = \sum_{k=1}^{\infty} \log( 1 - (\frac{x}{k\pi})^2 )
= - \sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{n}(\frac{x}{k\pi})^{2n}\tag{*}$$
Since the double series converges absolutely for $|x| < \pi$, we can switch the order of summation:
$$(*) = - \sum_{n=1}^{\infty}\frac{1}{n}\sum_{k=1}^{\infty}(\frac{x}{k\pi})^{2n} =
-\sum_{n=1}^{\infty}\frac{\zeta(2n)}{n}(\frac{x}{\pi})^{2n}\tag{**}
$$
We can also differentiate the result term by term to get:
$$\frac{\cos(x)}{\sin(x)} - \frac{1}{x} = -2 \sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}} x^{2n-1}$$
In the Taylor expansion of $\sin(x)$ and $\cos(x)$, their coefficients are all rational numbers. This mean when we expand L.H.S as a power series in $x$, all its coefficients are rational numbers too. Compare it with those in R.H.S, we get $\frac{\zeta(2n)}{\pi^{2n}}$ are rational numbers for all $n$.
A: Rewrite the integral as
$$\begin{align}\int_0^{\infty} dx \: x^{2 n+1} e^{-\pi x} \sum_{k=0}^{\infty} e^{- k \pi x} &= \sum_{k=0}^{\infty} \int_0^{\infty} dx \: x^{2 n+1} e^{-(k+1) \pi x}\\ &= \frac{(2 n+1)!}{\pi^{2 n+2}} \zeta(2 n+2)\end{align}$$
(in the last step, I evaluated the integral and the sum produced a Riemann zeta).  We know that, for a zeta of an even integer is 
$$\zeta(2 n+2) = (-1)^n \frac{B_{2 n+2} (2 \pi)^{2 n+2}}{2 (2 n+2)!}$$
where $B_k$ is the $k$-th Bernoulli number and is rational.  Putting the last two equations together, it is clear that the sought integral is rational for all natural numbers $n$.
