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#>



Deriving Newtonian capacity of sphere from Brownian motion
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](https://books.google.de/books?id=JnRIeuv3GJYC&pg=PA55&lpg=PA55)




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**[\*]** To see that the contribution with $r\_0\sin[u(r-r\_0)]$ in the integrand is indeed vanishing, use this [delta function identity:](https://en.wikipedia.org/wiki/Sokhotski-Plemelj_theorem)


$$\int\_0^\infty e^{ikx}\,dx=\pi\delta(k)+{\cal P}\frac{i}{k}\Rightarrow\int\_{r\_0}^\infty \sin[u(r-r\_0)]\,dr={\cal P}\frac{1}{u}, $$


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.


