TITLE: Average Value of Area Closed by Brownian Motion
QUESTION [5 upvotes]: Two dimensional brownian motion will intersect its own path infinitly many times. What is the average value of area, closed by curve during an intersection in brownian motion?

REPLY [6 votes]: There is a beautiful connection between the area swept out by Brownian motion and the Dirichlet function:
$L(s)=\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)^s}$.

Proposition: Let $A_t$ be the area swept out by the Brownian motion up to time $t$. Then for every $\alpha \ge 0$
  $$\mathbb{E}(|A_t|^\alpha)=\frac{4 \Gamma(1+\alpha)}{\pi^{1+\alpha}} L(1+\alpha)t^\alpha$$

Proof: Let $B_t=(B^1_t,B^2_t)$ be a two-dimensional Brownian motion. The algebraic area enclosed by $B$ up to time $t$ is given by 
$A_t=\frac{1}{2} \int_0^t B_s^1 dB_s^2-B_s^2dB_s^1$
where the integral is understood as a Ito integral. It is an easy exercise to check that
$A_t=\frac{1}{2} \beta_{\int_0^t \rho_s^2 ds}$    
where $\rho_t=\| B_t \|$ and $\beta$ is a Brownian motion independent from $\rho$.
Thus we have,
$\mathbb{E}(| A_t |^\alpha)=\frac{1}{2^\alpha} \mathbb{E}\left(\left| \beta_{\int_0^t \rho_s^2 ds} \right|^\alpha \right)$
Using the Brownian scaling property and the independence, we obtain
$\mathbb{E}(| A_t |^\alpha)=\frac{1}{2^\alpha} \mathbb{E}\left(| \beta_{1} |^\alpha\right) \mathbb{E}\left(\left( \int_0^t \rho_s^2 ds\right)^{\alpha/2}\right)=\frac{t^\alpha}{2^\alpha} \mathbb{E}\left(| \beta_{1} |^\alpha\right) \mathbb{E}\left(\left( \int_0^1 \rho_s^2 ds\right)^{\alpha/2}\right)$
Now, it is well-known that for every $\lambda \ge 0$
$\mathbb{E}\left(e^{-\frac{\lambda^2}{2} \int_0^1 \rho_s^2 ds}\right)=\frac{1}{\cosh \lambda} .$ 
We can deduce from this Laplace transform the following Mellin transform formula
$\mathbb{E}\left(\left( \int_0^1 \rho_u^2 du\right)^{s}\right)=\Gamma(1+s)2^{1+s} \left(\frac{2}{\pi} \right)^{2s+1}L(1+2s)$
The result follows then easily.