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?

Solution 1:

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

Solution 2:

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.