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
![\displaystyle \mathbb{E}[ \, \tau \, ] \; = \; \sum_{k \geq 1} |a_k|^2](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cmathbb%7BE%7D%5B+%5C%2C+%5Ctau+%5C%2C+%5D+%5C%3B+%3D+%5C%3B+%5Csum_%7Bk+%5Cgeq+1%7D+%7Ca_k%7C%5E2+&bg=ffffe3&fg=000000&s=0 "\displaystyle \mathbb{E}[ \, \tau \, ] \; = \; \sum_{k \geq 1} |a_k|^2")
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
![\displaystyle \mathbb{E}[ \, \|B_{\tau}\|^2 \, ] \; = \; 2 \mathbb{E}[\, \tau \, ].](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cmathbb%7BE%7D%5B+%5C%2C+%5C%7CB_%7B%5Ctau%7D%5C%7C%5E2+%5C%2C+%5D+%5C%3B+%3D+%5C%3B+2+%5Cmathbb%7BE%7D%5B%5C%2C+%5Ctau+%5C%2C+%5D.+&bg=ffffe3&fg=000000&s=0 "\displaystyle \mathbb{E}[ \, \|B_{\tau}\|^2 \, ] \; = \; 2 \mathbb{E}[\, \tau \, ].")
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
![\displaystyle \mathbb{E}[ \, \|B_{\tau}\|^2 \, ] = \mathbb{E}[ \, \|f(W_{\rho})\|^2 \, ] = \frac{1}{2\pi} \int_{\theta=0}^{2\pi} |f(e^{i\theta})|^2 \, d\theta.](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cmathbb%7BE%7D%5B+%5C%2C+%5C%7CB_%7B%5Ctau%7D%5C%7C%5E2+%5C%2C+%5D+%3D+%5Cmathbb%7BE%7D%5B+%5C%2C+%5C%7Cf%28W_%7B%5Crho%7D%29%5C%7C%5E2+%5C%2C+%5D+%3D+%5Cfrac%7B1%7D%7B2%5Cpi%7D+%5Cint_%7B%5Ctheta%3D0%7D%5E%7B2%5Cpi%7D+%7Cf%28e%5E%7Bi%5Ctheta%7D%29%7C%5E2+%5C%2C+d%5Ctheta.+&bg=ffffe3&fg=000000&s=0 "\displaystyle \mathbb{E}[ \, \|B_{\tau}\|^2 \, ] = \mathbb{E}[ \, \|f(W_{\rho})\|^2 \, ] = \frac{1}{2\pi} \int_{\theta=0}^{2\pi} |f(e^{i\theta})|^2 \, d\theta.")
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 ![{\mathbb{E}[ \, \tau \, ]}](http://s0.wp.com/latex.php?latex=%7B%5Cmathbb%7BE%7D%5B+%5C%2C+%5Ctau+%5C%2C+%5D%7D&bg=ffffe3&fg=000000&s=0 "{\mathbb{E}[ \, \tau \, ]}")
. 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).![E [||f(W_\rho)||^2]=\frac{1}{2\pi}\int_0^{2\pi}|f(e^{i.\theta}|^2d\theta](http://s0.wp.com/latex.php?latex=E+%5B%7C%7Cf%28W_%5Crho%29%7C%7C%5E2%5D%3D%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Cint_0%5E%7B2%5Cpi%7D%7Cf%28e%5E%7Bi.%5Ctheta%7D%7C%5E2d%5Ctheta&bg=ffffe3&fg=000000&s=0 "E [||f(W_\rho)||^2]=\frac{1}{2\pi}\int_0^{2\pi}|f(e^{i.\theta}|^2d\theta")
, 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 ?
By the way, when you say that
Alekk said,
December 28, 2012 at 5:59 pm
thanks!
And yes, that’s one way of proving that.