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.