# Zeta function for negative integers

I already proved that $\zeta(z)=\frac{1}{\Gamma(z)}\int_0^\infty\frac{t^{z-1}}{e^t-1}dt=\frac{\Gamma(z-1)}{2\pi i}\int_{-\infty}^0\frac{t^{z-1}}{e^{-t}-1}dt$

Now the Benoulli numbers are defined by $\frac{1}{e^t-1}=\sum_{m=0}^{\infty}B_m\frac{t^{m-1}}{m!}$ where $B_0=1, B_1=1/2, B_{2m+1}=0$

How can I use these things to get an expression for $\zeta(-n), n=0,1,2,3...$ in terms of $B_n$

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Have a look at Stopple "Primer in Analytic Number Theory" Section 8.2, especially the problems and solutions in the back. Or Edwards "Riemann Zeta Function" Section 1.5 – TheBirdistheWord Mar 11 '14 at 16:14

I'm assuming you mean $$\zeta(z) = \frac{\Gamma(1-z)}{2 \pi i} \int_{C} \frac{t^{z-1} }{e^{-t}-1} \ dt = \frac{\Gamma(1-z)}{2 \pi i} \int_{C} \frac{t^{z-1}e^{t} }{1-e^{t}} \ dt$$

where $C$ is a contour on the complex plane that starts at $- \infty$ below the branch cut on the negative real axis, goes around the origin, and then goes to back to $-\infty$ above the branch cut.

That representation of the zeta function is valid for all complex values of $z$ except $z=1$.

Now let $z= - n$.

Then $$\zeta(-n) = \frac{\Gamma(n+1)}{2 \pi i} \int_{C} \frac{t^{-(n+1)}e^{t}}{1-e^{t}} \ dt$$

But now since $z$ is an integer, the integral above and below the cut cancel each other. So all that is left is the circle around the origin.

So

$$\zeta(-n) = \frac{\Gamma(n+ 1)}{2 \pi i } \ 2 \pi i \ \text{Res}_{t=0} \left(\frac{t^{-n-1} e^{t}}{1-e^{t}} \right) = -n! \ \text{Res}_{t=0} \ \left( t^{-n-2} \frac{t e^{t}}{e^{t}-1} \right)$$

$$= - n! \ \text{Res}_{t=0} \left( t^{-n-2} \sum_{k=0}^{\infty} \frac{B_{m}(1)}{m!} t^{m} \right) = - n! \frac{B_{n+1}(1)}{(n+1)!} = - \frac{B_{n+1}}{n+1}$$

EDIT:

The above is only true for $n \ge 1$.

For $n=0$, $B_{1}(1) = \frac{1}{2} \ne B_{1}(0) = B_{1} =- \frac{1}{2}$.

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We have for $s$ with $\Re{e}(s) > 1$ :

$\Gamma(s)\zeta(s)=\int_0^\infty\frac{t^{s-1}}{{\rm e}^t-1}\mathrm dt=\int_0^1\frac{t^{s-1}}{\mathrm e^t-1}\mathrm dt+\int_1^\infty\frac{t^{s-1}}{\mathrm e^t-1} dt$.

The second integral is holomorphe in $s$. We take the Taylor series in the first integral.

We have for all $t$ with $|t| < 2\pi$

$\frac t{{\rm e}^t-1}=\sum_{n=0}^\infty\frac{B_n t^n}{n!}$,

where the $B_n$ are the Bernoulli's numbers. Integrating term by term

$\zeta(s)=\frac1{(s-1)\Gamma(s)}+\frac1{\Gamma(s)}\sum_{n=1}^\infty\frac{B_n}{n!(n+s-1)}+\frac1{\Gamma(s)}\int_1^\infty\frac{t^{s-1}}{\mathrm e^t-1} \mathrm dt$.

The series converge and is holomorphic for all s but $s=-n$, $(n \in \mathbb{N})$ because the convergence radius of the series is not modified by division by $n + s – 1$.

When $s \to k$, as $Γ(s)$ has a simple pole in $s=–k$ , $ζ(s)$ is the sum of one term which tend to $0$ and $\frac1{\Gamma(s)}~\frac{B_{k+1}}{(k+1)!~(s+k)}~\underset{\overset{s\to-k}{}}{\sim}~\frac{s+k}{\text{Res}(\Gamma,-k)}~\frac{B_{k+1}}{(k+1)!~(s+k)}=(-1)^k~k!~\frac{B_{k+1}}{(k+1)!}.$

We have the Euler formula : $\zeta(-k)=(-1)^k\frac{B_{k+1}}{k+1}.$

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