# Analytic Continuation of the Riemann Zeta Function

The Riemann Zeta function is defined as the series

$$\zeta(s)=\sum_{n=1}^{+\infty} \frac{1}{n^s} \ \ \ \ \ \ \ \ \ \ (1)$$

which converges for $$\Re (s)>1$$.

It can be defined by analytic continuation to the whole complex plane except $$s=1$$ through the Riemann functional equation

$$\zeta(s)=2^s \pi^{s-1} \sin \left({\pi s \over 2} \right) \Gamma(1-s) \zeta(1-s) \ \ \ \ \ \ \ \ \ \ (2)$$

where $$\Gamma(s)$$ is the gamma function.

We want to derive the integral equation and obtain the values of the $$\zeta$$ function for negative integers

$$\zeta(-n)=-{B_{n+1}\over n+1} ,\ \ \ \ \ \ \ \ \ \ (3)$$

and in particular the value

$$\zeta(-1)=\sum_{n=1}^{+\infty} n=-{1\over 12}.\ \ \ \ \ \ \ \ \ \ (4)$$

# Using smoothed sums

Terry Tao gives a beautiful interpretation to the dry analytic continuation result using smoothed sums. It shows, for example, that

$$\sum_{n=1}^{+\infty} \eta(n/N) = -{1\over 2} + C_{\eta,0} N + O(\frac{1}{N}) \ \ \ \ \ \ \ \ \ \ (5)$$

where $$\eta(x)$$ is a smooth cutoff function with support in $$[0,1]$$ and $$C_{\eta,s}$$ is the Mellin transform of the cutoff function

$$C_{\eta,s}=\int_0^\infty x^s\eta(x)dx \ \ \ \ \ \ \ \ \ \ (6)$$

This expression, $$(5)$$, shows that the result of an analytic continuation of a divergent sum like this gives the unique non-divergent part of the series.