Zeta function for nonpositive 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 assume you mean $$\zeta(z) =  \frac{\Gamma(1-z)}{2 \pi i} \int_{C} \frac{t^{z-1} }{e^{-t}-1} \, \mathrm dt = -\frac{\Gamma(1-z)}{2 \pi i} \int_{C} \frac{t^{z-1}e^{t} }{e^{t}-1} \, \mathrm dt,$$
where $C$ is a contour that starts at $- \infty$ below the branch cut on the negative real axis, goes around the origin (without enclosing any of the points $z= \pm 2  \pi i, \pm 4 \pi i, \ldots$), and then goes to back to $-\infty$ above the branch cut.
The above integral representation of the Riemann zeta function is valid for all complex values of $z$ except positive integers.
If we let $z= - n$, where $n \in \mathbb{N}_{\ge 0}$, we get
$$ \zeta(-n) = -\frac{\Gamma(n+1)}{2 \pi i} \int_{C} \frac{t^{-n-1}e^{t}}{e^{t}-1} \, \mathrm  dt.$$
But since $z$ is now an integer, the integral above and below the branch cut cancel each other, and all we're left with is the circle around the origin.
Therefore,
$$ \begin{align}  \zeta(-n) &=  -\frac{\Gamma(n+ 1)}{2 \pi i } \, 2 \pi i \, \operatorname{Res}_{t=0} \left(\frac{t^{-n-1} e^{t}}{e^{t}-1}  \right) \\ &= - n! \, \operatorname{Res}_{t=0} \left( t^{-n-1}  \sum_{m=0}^{\infty} \frac{B_{m}(1)}{m!} t^{m-1} \right) \tag{1}\\ & = - n! \, \frac{B_{n+1}(1)}{(n+1)!} \\ &= - \frac{B_{n+1}(1)}{n+1} \\&= (-1)^{n} \frac{B_{n+1}}{n+1}. \end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Bernoulli_polynomials#Generating_functions
A: $\newcommand{\+}{^{\dagger}}
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$$
\mbox{Use}\quad\zeta\pars{-n}=
-2^{-n}\pi^{-n - 1}\Gamma\pars{1 + n}\zeta\pars{1 + n}\sin\pars{n\pi \over 2}
$$
A: 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 :
$\zeta(-k)=(-1)^k\frac{B_{k+1}}{k+1}.$
