Well first, the Riemann zeta function for the first negative odd integers:
$\zeta(-1) = -\frac{1}{12}$
$\zeta(-3) = \frac{1}{120}$
$\zeta(-5) = -\frac{1}{252}$
$\zeta(-7) = \frac{1}{240}$
and so on….
Why are these rational? I have tried searching Google and browsing through other questions but I don't know of any proof of that. My first thought is that it would be very complicated (maybe a proof by contradiction?) but I still do not know of any proof of this.
My question is if there is a nice proof for why this all works?
Best Answer
For $\Re(s) > 1$ and by analytic continuation for $\Re(s) > -K$ $$\zeta(s) \Gamma(s) = \int_0^\infty \frac{x^{s-1}}{e^x-1}dx= \sum_{k=0}^K\frac{B_k}{k!} \frac{1}{s+k-1}+\int_0^\infty x^{s-2}(\frac{x}{e^x-1}-\sum_{k=0}^K\frac{B_k}{k!} x^k 1_{x < 1}) dx$$ ie. the Taylor expansion of $\frac{x}{e^x-1}$ gives the residues of the simple poles of $\zeta(s)\Gamma(s)$ at negative integers. They are rational numbers because the Taylor series for $\frac{e^x-1}{x}$ has rational coefficients.
For other Dirichlet series we have
$$\sum_{n=1}^\infty a_n n^{-s} \Gamma(s) = \int_0^\infty x^{s-1}\sum_{n=1}^\infty a_n e^{-nx}dx$$ but in general $x^m\sum_{n=1}^\infty a_n e^{-nx}$ has no reason to be smooth at $x=0$ nor to have rational derivatives.