## Let gooby do teh hoemwerk: Riemann zeta at two

Based on what I read in a very old book, we show

$\zeta(2) = \sum\limits_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$

We can divide the sum into even and odd contributions

$\sum\limits_{n=1}^\infty \frac{1}{n^2} = \sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2} + \sum\limits_{n=1}^\infty \frac{1}{(2n)^2}$

One sees that

$\zeta(2) =\frac{4}{3}\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^2}$

In order to evaluate the right hand side of this equation we consider the function

$C(x) = \sum\limits_{n=1}^\infty \frac{\cos (2n-1)x}{(2n-1)^2}$, such that $\zeta(2) =\frac{4}{3}C(0)$

We now take a small detour and consider the finite sum $C_N(x)$ and its derivatives

$C_N(x) = \sum\limits_{n=1}^N \frac{\cos (2n-1)x}{(2n-1)^2}$

$C_N'(x) = -\sum\limits_{n=1}^N \frac{\sin (2n-1)x}{2n-1}$

$C_N''(x) = -\sum\limits_{n=1}^N \cos(2n-1)x$

Multiplying the last equation by $2\sin x$ and using

$2\sin x \cos k x = \sin(k+1)x-\sin(k-1)x$

We get

$2\sin x \,C_N''(x) = -\sin 2Nx$

Now one can solve for $C_N'(x)$

$C_N'(x) = C_N'(\pi/2) - \int\limits_{\pi/2}^{x}dy\,\frac{\sin 2Ny}{\sin y}$

By writing

$C_N'(x) = C_N'(\pi/2) + \frac{1}{2N} \int\limits_{\pi/2}^{x}dy\,\frac{\frac{d}{dy}\cos 2Ny}{\sin y}$

One can think of integration by parts and notice that the second term vanishes when $N\to\infty$. Now

$C'(\pi/2) = -\lim\limits_{N\to\infty}\sum\limits_{n=1}^N \frac{(-1)^n}{2n-1} =-\arctan(1) = -\frac{\pi}{4}$

In other words,

$C'(x) =-\frac{\pi}{4}$

from which it is immediate

$C(x) = C(0) - \frac{\pi}{4}x$

Now note $C(\pi/2) = 0$ and therefore

$C(0) = \frac{\pi}{4}\times\frac{\pi}{2} = \frac{\pi^2}{8}$

Finally, the identity $\zeta(2) =\frac{4}{3}C(0)$ establishes the result $\zeta(2) =\frac{\pi^2}{6}$.

Happy new year to evri1!