What is the power series expansion for Riemann-Zeta at $0$? What are the first few terms of the Laurent series expansion of $\zeta(0)$? It gets mentioned here but they only show the first term and I am kind of confused on how they got $-1/2$.
 A: We can use the integral, which for $x\gt1$, can be seen to be
$$
\begin{align}
\int_0^\infty\frac{xt^{x-1}}{e^t+1}\mathrm{d}t
&=\int_0^\infty\frac{xt^{x-1}}{1+e^{-t}}e^{-t}\;\mathrm{d}t\\
&=x\sum_{k=1}^\infty(-1)^{k-1}\int_0^\infty t^{x-1}e^{-kt}\;\mathrm{d}t\\
&=x\sum_{k=1}^\infty(-1)^{k-1}k^{-x}\int_0^\infty t^{x-1}e^{-t}\;\mathrm{d}t\\[6pt]
&=x\eta(x)\Gamma(x)\\[12pt]
&=(1-2^{1-x})\zeta(x)\Gamma(x+1)\tag{1}
\end{align}
$$
Since $(1)$ is an analytic function of $x$, and we can integrate by parts to get
$$
\begin{align}
\lim_{x\to0^+}\int_0^\infty\frac{xt^{x-1}}{e^t+1}\mathrm{d}t
&=\lim_{x\to0^+}\int_0^\infty\frac{t^xe^t}{(e^t+1)^2}\mathrm{d}t\\
&=\int_1^\infty\frac{\mathrm{d}u}{(u+1)^2}\\
&=\frac{1}{2}\tag{2}
\end{align}
$$
not only do we see that the integral in $(2)$ converges for $x\gt-1$, but also, by comparing with $(1)$, that for $x=0$, we get by analytic continuation that
$$
\zeta(0)=-\frac12\tag{3}
$$
A: A globally convergent series for the Riemann Zeta except $s=1
 $ is $$\zeta\left(s\right)=\frac{1}{1-2^{1-s}}\sum_{n\geq0}\frac{1}{2^{n+1}}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(k+1\right)^{-s}
 $$ hence $$\zeta\left(0\right)=-\sum_{n\geq0}\frac{1}{2^{n+1}}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}=-\frac{1}{2}-\sum_{n\geq1}\frac{1}{2^{n+1}}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}=-\frac{1}{2}
 $$ by binomial theorem.
