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
$$ \lim_{t\to \infty}\frac{1}{t}\int_{r_0}^\infty \,dr\int_{0}^{\infty}du\, \,u^{-1}\left(1- e^{-u^{2}t/2}\right)\sin[u(r-r_0)]=$$$$8r_0^2 \lim_{t\to \infty}\frac{1}{t}\int_{r_0}^\infty \,dr\int_{0}^{\infty}du\, \,u^{-1}\left(1- e^{-u^{2}t/2}\right)\sin[u(r-r_0)]=$$
$$=\lim_{t\to \infty}\frac{1}{t}\int_{0}^{\infty}\,u^{-2}\left(1- e^{-u^{2}t/2}\right)du=\lim_{t\to \infty}\sqrt\frac{\pi}{2t}=0,$$$$=8r_0^2\lim_{t\to \infty}\frac{1}{t}\int_{0}^{\infty}\,u^{-2}\left(1- e^{-u^{2}t/2}\right)du=8r_0^2\lim_{t\to \infty}\sqrt\frac{\pi}{2t}=0,$$
and indeed, this contribution vanishes.