Q:

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 $B_n$

A:

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

A:

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$$
\mbox{Use}\quad\zeta\pars{-n}=
-2^{-n}\pi^{-n - 1}\Gamma\pars{1 + n}\zeta\pars{1 + n}\sin\pars{n\pi \over 2}
$$

