Here is a delightful probabilistic proof that extracted from a recent article by Greg Markowsky. This starts with pretty lemma (which can originally be found in this article) on the exit time of a dimensional Brownian motion . Suppose that is an analytic function on the neighbourhood of the unit disk. This function maps the unit disk to with boundary where . A two dimensional brownian motion started at takes on average

to exit the domain where and is the hitting time of the boundary . Indeed, since the situation is invariant by translation one can always suppose that . The proof of this result is a simple martingale argument. Indeed, since is a martingale, the optional stopping theorem shows that

Then, since is an analytic function, Ito’s formula shows that the trajectories of are the same as the trajectories of , up to a time change, where is another dimensional Brownian motion started at . Consequently, where is the hitting time . This shows that

Since Parseval‘s theorem (or a direct calculation) shows that the last quantity also equals , the conclusion follows.

To find a nice identity, it thus suffices to find an easy domain where one can compute explicitly the brownian exit time . The first thing that comes to mind is a strip since it is classical that a Brownian motion started at takes on average to exit the strip . Since maps the unit disk to the strip it follows that

which is indeed equivalent to the celebrated identity .

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## TheBridge said,

December 28, 2012 at 12:51 pm

Hi, nice and original proof of zeta(2).

By the way, when you say that , you are using the argument that rotation of a 2d BM is invariant in distribtuion so it has to be uniform on the circle, right ?

## Alekk said,

December 28, 2012 at 5:59 pm

thanks!

And yes, that’s one way of proving that.