## Let gooby do teh hoemwerk: Riemann zeta at minus one

Consider the problem: evaluate the Riemann zeta function $\zeta(-1)$. Reminder: for $\text{Re}z>1$ the zeta function is defined

$\zeta(z) = \sum\limits_{n=1}^\infty \frac{1}{n^z}$

In the rest of the complex plane the zeta function is defined as the analytic continuation (wherever possible). Clearly, we can’t simply put $-1$ in the above formula. We need to think of something to do the analytic continuation. The starting point is an integral representation…

$\zeta(z) = \sum\limits_{n=0}^\infty \frac{1}{(n+1)^z} = \sum\limits_{n=0}^\infty \frac{1}{n+1}\frac{1}{(n+1)^{z-1}} = \sum\limits_{n=0}^\infty \frac{1}{n+1}\frac{1}{(n+1)^{z-1}}\frac{1}{\Gamma(z)}\int\limits_0^\infty dy\,y^{z-1}e^{-y}$,

where I inserted unity by means of the Gamma function. We can work further on this

$\zeta(z) = \frac{1}{\Gamma(z)}\sum\limits_{n=0}^\infty\int\limits_0^\infty \frac{dy}{n+1}\Bigl(\frac{y}{n+1}\Bigr)^{z-1}e^{-y} = \frac{1}{\Gamma(z)}\sum\limits_{n=0}^\infty\int\limits_0^\infty dx\,x^{z-1}e^{-(n+1)x}$

where I substituted $y=x/(n+1)$. By using the geometric series one gets

$\zeta(z)= \frac{1}{\Gamma(z)}\int\limits_0^\infty dx\,\frac{x^{z-1}}{e^x-1}$.

We can make progress by splitting this integral

$\zeta(z)= \frac{1}{\Gamma(z)}\int\limits_0^1 dx\,\frac{x^{z-1}}{e^x-1} + \frac{1}{\Gamma(z)}\int\limits_1^\infty dx\,\frac{x^{z-1}}{e^x-1} \equiv \frac{1}{\Gamma(z)}\int\limits_0^1 dx\,\frac{x^{z-1}}{e^x-1} + \frac{S(z)}{\Gamma(z)}$,

where

$S(z)= \int\limits_1^\infty dx\,\frac{x^{z-1}}{e^x-1}$

is an entire function. We subtract it from the zeta function and make a Taylor expansion in the denominator:

$\zeta(z)-\frac{S(z)}{\Gamma(z)}= \frac{1}{\Gamma(z)}\int\limits_0^1 dx\,\frac{x^{z-1}}{x+\frac{x^2}{2}+\frac{x^3}{6}+\dots}$

and then expand the denominator (which is valid in the interval of integration)

$\zeta(z)-\frac{S(z)}{\Gamma(z)}= \frac{1}{\Gamma(z)}\int\limits_0^1 dx\,x^{z-2}\Bigl[1-\Bigl(\frac{x}{2}+\frac{x^2}{6}\Bigr)+\Bigl(\frac{x}{2}+\frac{x^2}{6}\Bigr)^2+\dots\Bigr]$

Now integrate term by term

$\zeta(z)-\frac{S(z)}{\Gamma(z)}= \frac{1}{\Gamma(z)}\Bigl[\frac{1}{z-1}-\frac{1}{2z}+\frac{1}{12(z+1)}+\dots\Bigr]$

Now we’re in business! Note that $\frac{1}{\Gamma(z)}$ is an entire function with zeroes at $z=0,-1,-2, \dots$. That means that for the point of interest the term proportional to $S(z)$ vanishes and with the help of

$\Gamma(z) = \frac{\Gamma(z+2)}{z(z+1)}$

we can take the limit

$\zeta(-1)= \lim\limits_{z\to -1}\frac{1}{\Gamma(z+2)}\Bigl[\frac{z(z+1)}{z-1}-\frac{z+1}{2}+\frac{z}{12}+(z+1)\times \dots\Bigr]$

which yields

$\zeta(-1) = -\frac{1}{12}$

We also see that the other terms indicated by the ellipses don’t contribute.