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.
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1.
Hardy, G. H. Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work. (AMS Chelsea Pub., 1999).