How to find $\zeta(0)=\frac{-1}{2}$ by definition? I would like to know how we can find the following result:
$$\zeta(0)=-\frac12$$
Is there a way, using the definition, $$\zeta(s)=\sum_{i=1}^{\infty}i^{-s}$$
to find this?
 A: Consider the integral
$$
\begin{align}
\int_0^\infty\frac{xt^{x-1}}{e^t+1}\mathrm{d}t
&=\int_0^\infty\frac{xt^{x-1}}{1+e^{-t}}e^{-t}\;\mathrm{d}t\\
&=x\sum_{k=1}^\infty(-1)^{k-1}\int_0^\infty t^{x-1}e^{-kt}\;\mathrm{d}t\\
&=x\sum_{k=1}^\infty(-1)^{k-1}k^{-x}\int_0^\infty t^{x-1}e^{-t}\;\mathrm{d}t\\
&=x\eta(x)\Gamma(x)\\
&=(1-2^{1-x})\zeta(x)\Gamma(x+1)\tag{1}
\end{align}
$$
Integrate by parts to get
$$
\begin{align}
\lim_{x\to0^+}\int_0^\infty\frac{xt^{x-1}}{e^t+1}\mathrm{d}t
&=\lim_{x\to0^+}\int_0^\infty\frac{t^xe^t}{(e^t+1)^2}\mathrm{d}t\\
&=\int_1^\infty\frac{\mathrm{d}u}{(u+1)^2}\\
&=\frac{1}{2}\tag{2}
\end{align}
$$
Sending $x$ to $0$ in $(1)$ and combining with $(2)$, we get $\zeta(0)=-\frac{1}{2}$.
A: The "defining sum" converges only for $\Re s > 1$. One can however use the related formula
$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty \frac1{2^{n+1}}\sum_{k=0}^n (-1)^k \binom{n}{k} (k+1)^{-s}$$
for $s=0$. This complicated sum can be derived by applying the Euler transformation to the series for Dirichlet $\eta$, the alternating version of Riemann's function.
