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)},


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.


About goobypl5

pizza baker, autodidact, particle physicist
This entry was posted in Math and tagged , , . Bookmark the permalink.

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