How does one derive this formula $\zeta(-n) =\frac{B_{n+1}}{n+1}$?

Question

$$\zeta(-n) =-\frac{B_{n+1}}{n+1}$$

What is the motivation or the derivation of this formula?

Answer 1

Use the integral

$$(e^{2\pi i s}-1)\zeta(s)\Gamma(s) = \int_C B(t)t^s\frac{dt}{t}$$

where $C$ is the keyhole contour and $B(t) = \frac{e^{-t}}{1-e^{-t}}$ is the exponential generating function of the Bernoulli numbers. This integral gives the analytic continuation of the Riemann zeta function. At negative integers, the two cuts of the contour collapse and you get a simple circular contour around the origin, which picks up the coefficients of $B(t)$ by the residue theorem.

This formula is important because it expresses the zeta function as the Mellin transform of the so-called Bernoulli distribution. By "normalizing" the Bernoulli distribution at a prime $p$, one gets a $p$-adic distribution whose $p$-adic Mellin transform is the $p$-adic Riemann zeta function constructed by Kubota and Leopoldt.

Answer 2

Use the functional equation $$\zeta(-s) = -\dfrac{\sin(\pi s/2)}{2^s \pi^{s+1}} \Gamma(1+s) \zeta(1+s)$$ From the function equation, you have $$\zeta(-n) = -\dfrac{B_{n+1}}{n+1}$$ It is worth noting that if $n$ is even, then $\zeta(-n) = 0$.