# Abstract Nonsense

## A Few More Sums Involving the Zeta Function

Point of post: In this post we compute the value of

$\displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{n(n+1)}$

The reason we do this particular sum is that while the mathworld.com site has many identities relating to the zeta function even a lot that look like this sum, they don’t have this one. It is slightly larder than the one’s on the site, so maybe this is the reason.

Problem: Compute

$\displaystyle S=\sum_{m=1}^{\infty}\frac{\zeta(2m)}{m(m+1)}$

Solution: Recall that for all $z\in\mathbb{C}$ one has

$\displaystyle \sin(\pi z)=\pi z\prod_{n=1}^{\infty}\left(1+\frac{z^2}{n^2}\right)$

And thus for $|z|<1$ we have that

\displaystyle \begin{aligned}\log\left(\sin\left(\pi\sqrt{z}\right)\right)-\log\left(\pi\sqrt{z}\right) &=\sum_{n=1}^{\infty}\log\left(1-\frac{z}{n^2}\right)\\ &= -\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{z^m}{m n^{2m}}\\ &= -\sum_{m=1}^{\infty}\frac{z^m}{m}\sum_{n=1}^{\infty}\frac{1}{n^{2m}}\\ &= -\sum_{m=1}^{\infty}\frac{\zeta(2m)}{m}z^m\end{aligned}

And thus, for $|x|<1$ we may conclude that

$\displaystyle \sum_{m=1}^{\infty}\frac{\zeta(2m)}{m(m+1)}x^{m+1}=\int_0^x \left(\log\left(\pi\sqrt{z}\right)-\log\left(\sin\left(\pi\sqrt{z}\right)\right)\right)\text{ }dz$

But, taking the limit as $x\to 1^{-}$ on both sides gives

$\displaystyle \sum_{m=1}^{\infty}\frac{\zeta(2m)}{m(m+1)}=\int_0^1 \left(\log\left(\pi\sqrt{z}\right)-\log\left(\sin\left(\pi\sqrt{z}\right)\right)\right)\text{ }dz$

where we’ve used the fact that the series on the left had radius of convergence $1$ and the series converged at $1$ and thus Abel’s Limit Theorem applied. Note then that

$\displaystyle S=\log(\pi)-\frac{1}{2}-I$

where

$\displaystyle I=\int_0^1 \log\left(\sin\left(\pi\sqrt{z}\right)\right)\text{ }dz$

Letting $\pi\sqrt{z}=t$ in this integral gives

$\displaystyle I=\frac{2}{\pi^2}\int_0^\pi t \log\left(\sin\left(t\right)\right)\text{ }dt$

Let $\displaystyle t=\pi-y$ in this integral to get that

$\displaystyle I=\frac{2}{\pi^2}\int_0^\pi \left(\pi-y\right)\log\left(\sin\left(y\right)\right)\text{ }dy=\frac{2}{\pi^2}\int_0^\pi \log\left(\sin\left(y\right)\right)\text{ }dy-I$

and thus solving we find that

$\displaystyle \pi I=\int_0^\pi \log\left(\sin(y)\right)\text{ }dy$

Letting $\displaystyle u=y-\frac{\pi}{2}$ gives us that

$\displaystyle \pi I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\log\left(\cos(u)\right)\text{ }du$

Note though that $\cos(u)$ is symmetric on $\displaystyle \left[\frac{-\pi}{2},\frac{\pi}{2}\right]$ and thus

$\displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\cos(t)\right)\text{ }dt$

or letting $\displaystyle w=\frac{\pi}{2}-u$ gives us

$\displaystyle \frac{\pi}{2}I=\int_0^{\frac{\pi}{2}}\log\left(\sin(w)\right)\text{ }dw$

Thus,

$\displaystyle \pi I=\int_0^{\frac{\pi}{2}}\left(\log\left(\cos(w)\right)+\log\left(\sin(w)\right)\right)\text{ }dw=\frac{-\pi\log(2)}{2}+\int_0^{\frac{\pi}{2}}\log\left(\sin(2w)\right)\text{ }dw$

where we’ve used the fact that $\displaystyle \sin(w)\cos(w)=\frac{1}{2}\sin(2w)$. Let $2w=x$ in this last integral to get that

$\displaystyle \pi I=\frac{-\pi\log(2)}{2}+\frac{1}{2}\int_0^{\pi}\log\left(\sin(x)\right)\text{ }dx=\frac{-\pi\log(2)}{2}+\frac{pi}{2}I$

Solving we find that

$\displaystyle I=-\log(2)$

and thus

$\displaystyle S=\log(\pi)-\frac{1}{2}-I=\log\left(2\pi\right)-\frac{1}{2}$

References:

1. Andrews, George E., Richard Askey, and Ranjan Roy. Special Functions. Cambridge [u.a.: Cambridge Univ., 2006. Print.