Let $f(x,y,z)=(x^2+y^2+z^2)^{-\frac12}$
Let $B=(B^1,B^2,B^3)$ be a Brownian motion with value in $\mathbb R^3$
I want to compute $E[f(B_t)]$
In my note I see that this correspond to $$\int_{\mathbb R^3}\frac1{x_1^2+x_2^2+x_3^2}\frac1{(2\pi t)^\frac32}e^{-\frac{x_1^2+x_2^2+x_3^2}{2t}}$$
I'm not able to understand why the density have to be the product of the three normal distributed random variable (that in this case are the B.M.). I know that they are independent, but still it is not clear to me.
If $X_1,X_2,...,X_n$ are independent random variables having densities $f_1,f_2,...,f_n$ then the joint density function of $(X_1,X_2,...,X_n)$ is $f_1(x_1)f_2(x_2)....f_n(x_n)$.
To evaluate the integral use spherical coordinates: http://mathworld.wolfram.com/SphericalCoordinates.html
The evaluation is very simple once you change to spherical coordinates.