Deriving Newtonian capacity of sphere from Brownian motion We have the following result by Spitzer (see (1) or Port)
$lim_{t\to \infty}\frac{1}{t}\int_{\mathbb{R}^{n}/B_{r_{0}}}P_{x}(T_{B_{r_{0}}}<t)dx=Cap(B_{r_{0}})=\frac{r_{0}}{4\pi}$
By Chuancun and Rong (2) we have
$P_{x}(T_{B_{r_{0}}}<t)=\frac{2r_{0}}{\pi |x|}\int_{0}^{\infty}u^{-1}(1-e^{\frac{-u^{2}t}{2}})sin(u(|x|-r_{0}))du.$
Therefore, $\frac{r_{0}}{4\pi}=Cap(B_{r_{0}})=lim_{t\to \infty}\frac{1}{t}\int_{\mathbb{R}^{n}/B_{r_{0}}}\frac{2r_{0}}{\pi |x|}\int_{0}^{\infty}u^{-1}(1-e^{\frac{-u^{2}t}{2}})sin(u(|x|-r_{0}))dudx$
I am having trouble going from the integral expression to $\frac{r_{0}}{4\pi}$. Has anyone done this before?
Attempt
One can simplify the inner integral to 
$lim_{t\to \infty}\frac{1}{t}\int_{\mathbb{R}^{n}/B_{r_{0}}}\frac{2r_{0}}{\pi |x|}[\frac{\pi}{2}-\frac{\pi}{2}erf(\frac{|x|-r_{0}}{\sqrt{2t}})]dx=$
$4\pi r_{0} lim_{t\to \infty}\frac{1}{t}\int_{r_{0}}^{\infty}[1-\frac{2}{\sqrt{\pi}}\int_{0}^{\frac{r-r_{0}}{2t}}e^{-u^{2}}du]r dr=$
$= \frac{r_{0}}{4\pi}$
which means that we must have
$lim_{t\to \infty}\frac{1}{t}\int_{r_{0}}^{\infty}[1-\frac{2}{\sqrt{\pi}}\int_{0}^{\frac{r-r_{0}}{2t}}e^{-u^{2}}du]r dr=\frac{1}{(4\pi)^{2}}$.
I find this hard to believe since there is an arbitrary $r_{0}$ on LHS but not on RHS. I will type as find things. If you have a solution or ideas please post.
thank you
(1)http://www.math.utah.edu/~mendez/capacities.pdf
(2)http://math.scichina.com:8081/sciAe/EN/abstract/abstract377404.shtml#
 A: First of all, please note that your second equation should have $u^{-1}$ instead of $u$, see page 580 of the Chuancun & Rong paper you cite. So we need to evaluate (in $n=3$ dimensions):
$$Cap(B_{r_{0}})=\lim_{t\to \infty}\frac{1}{t}\int_{r_0}^\infty 4\pi r^2\,dr\int_{0}^{\infty}du\, \frac{2r_0}{\pi r}\,u^{-1}\left(1- e^{-u^{2}t/2}\right)\sin[u(r-r_0)]$$
we have the following two identies:
$$\int_{r_0}^\infty (r-r_0) \sin[u(r-r_0)]\,dr=-\frac{d}{du}\int_{0}^\infty \cos ur\,dr=-\pi\frac{d}{du}\delta(u)$$
$$\int_0^\infty u^{-1}\left(1-e^{-u^2 t/2}\right)\frac{d}{du}\delta(u)\,du=\int_0^\infty (ut/2)\frac{d}{du}\delta(u)\,du=-t/4$$
the definition of the capacitance contains $r\sin[u(r-r_0)]$ instead of $(r-r_0) \sin[u(r-r_0)]$, but the difference gives a vanishing contribution [*].
putting everything together I arrive at
$$Cap(B_{r_{0}})=2\pi r_0$$
 instead of the $r_0/4\pi$ you were looking for, but actually I think $2\pi r_0$ is the correct answer, see for example page 111 of Random Walks, Brownian Motion, and Interacting Particle Systems

[*] To see that the contribution with $r_0\sin[u(r-r_0)]$ in the integrand is indeed vanishing, use this delta function identity:
$$\int_0^\infty e^{ikx}\,dx=\pi\delta(k)+{\cal P}\frac{i}{k}\Rightarrow\int_{r_0}^\infty \sin[u(r-r_0)]\,dr={\cal P}\frac{1}{u}, $$
where ${\cal P}$ indicates the Cauchy principal value. So the contribution in question is 
$$8r_0^2 \lim_{t\to \infty}\frac{1}{t}\int_{r_0}^\infty \,dr\int_{0}^{\infty}du\, \,u^{-1}\left(1- e^{-u^{2}t/2}\right)\sin[u(r-r_0)]=$$
$$=8r_0^2\lim_{t\to \infty}\frac{1}{t}\int_{0}^{\infty}\,u^{-2}\left(1- e^{-u^{2}t/2}\right)du=8r_0^2\lim_{t\to \infty}\sqrt\frac{\pi}{2t}=0,$$
and indeed, this contribution vanishes.
