TITLE: Prove that $\int_0^{\infty} \frac{x^{2n+1}}{e^{\pi x}-1}\mathrm{d}x \in \mathbb{Q}, \forall \,n\in\mathbb{N}$
QUESTION [5 upvotes]: Prove that
$$\int_0^{\infty} \frac{x^{2n+1}}{e^{\pi x}-1}\mathrm{d}x \in \mathbb{Q}, \forall \,n\in\mathbb{N}$$

REPLY [3 votes]: 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 $n$.