sequences and series - 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?

Answer

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 $\zeta(0)=-\frac{1}{2}$.