Abstract Nonsense

Crushing one theorem at a time

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


\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


\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}



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


January 4, 2011 - Posted by | Analysis, Computations, Fun Problems | , , , ,

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