Integration of messy expression I would like to know the following integral:
$$
\int\mathrm{d}^{3}x'\,{z' - v t \over
\,\sqrt{\,\left(x - x'\right)^{2} + \left(y - y'\right)^{2} +
\left(z - z'\right)^{2}\,}\,}\,
{1 \over \left[x'^{\,2} + y\,'^{\,2} + \left(z'- vt\right)^2\right]^{3/2}}
$$
Background: the integral is a result of a convolution of the Green's function for the Laplacian in $\mathbb{R}^{3}$ and a source term proportional to
$\dfrac{z' - v t}{\left[x'^{\,2} + y\,'^{\,2} + \left(z'- vt\right)^{2}\right]^{3/2}}$.
 A: Let $\vec{u} = (x,y,z-vt)$, $\vec{u}' = (x',y',z'-vt)$ and parametrize them using
polar coordinates,
$$\begin{align}
\vec{u}  &= (x,y,z-vt) = (r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta)\\
\vec{u}' &= (x',y',z'-vt) = (r'\sin\theta'\cos\phi',r\sin\theta'\sin\phi',r\cos\theta')
\end{align}
$$
In terms of these polar coordinates, we have following multipole expansion:
$$\frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}}
= \frac{1}{|\vec{u} - \vec{u'}|}
= \sum_{\ell=0}^\infty \frac{r_<^{\ell}}{r_>^{\ell+1}}P_\ell(\cos\gamma)
$$
where 
$\begin{cases} 
r_< &= \min\{ r, r' \}\\
r_> &= \max\{ r, r' \}
\end{cases}
$, $\gamma$ is the angle between $\vec{u}$ and $\vec{u'}$ and
$P_\ell(x)$ are Legendre polynomials.
Let $\Omega'$ be the unit sphere for $\vec{u'}$ and $d\Omega' = \sin\theta' d\theta'd\phi'$ be its area element.
We can rewrite the integral at hand as
$$
\int_0^\infty \int_{\Omega'} \left(\sum_{\ell=0}^\infty 
\frac{r_<^\ell}{r_>^{\ell+1}}P_\ell(\cos\gamma)\right) \frac{r'\cos\theta'}{r'^3}r'^2 dr' d\Omega'
=  \sum_{\ell=0}^\infty
\left(\int_0^\infty \frac{r_<^\ell}{r_>^{\ell+1}} dr'\right)
\left(\int_{\Omega'} P_\ell(\cos\gamma)\cos\theta' d\Omega'\right)
$$
We can decompose Legendre polynomials into spherical harmonics:
$$\begin{align}
P_\ell(\cos\theta') &= \sqrt{\frac{4\pi}{2\ell+1}}Y_{\ell 0}(\theta',\phi')\\
P_\ell(\cos\gamma)  &= \frac{4\pi}{2\ell+1}\sum_{m=-\ell}^\ell Y_{\ell m}(\theta,\phi) Y_{\ell m}^*(\theta',\phi')
\end{align}
$$
Since spherical harmonics are orthonormal,
$$\int_{\theta'=0}^\pi \int_{\phi'=0}^{2\pi} Y_{\ell m}(\theta',\phi')Y_{\ell' m'}^*(\theta', \phi') d\Omega' = \delta_{\ell\ell'} \delta_{mm'}$$
We find
$$\begin{align}
  \int_{\Omega'} P_\ell(\cos\gamma)\cos\theta' d\Omega'
= & \int_{\Omega'} P_\ell(\cos\gamma)P_1(\cos\theta') d\Omega'\\
= & \int_{\Omega'} \left(\frac{4\pi}{2\ell+1}\sum_{m=-\ell}^{m=\ell}Y_{\ell m}(\theta,\phi)Y_{\ell m}^*(\theta',\phi')\right)\left(\sqrt{\frac{4\pi}{3}}Y_{10}(\theta',\phi')\right) d\Omega'\\
= & \frac{4\pi}{2\ell+1}\delta_{\ell 1} \times 
\sqrt{\frac{4\pi}{3}} Y_{1 0}(\theta,\phi)\\
= & \frac{4\pi}{3}\cos\theta\, \delta_{\ell 1}
\end{align}
$$
This allow us to simplify the integral to
$$\frac{4\pi}{3} \cos\theta \int_0^\infty \frac{r_<}{r_>^2} dr'
= \frac{4\pi}{3}\cos\theta \left[
\int_0^r \frac{r'}{r^2} dr' + \int_r^\infty \frac{r}{r'^2} dr' \right]
= 2\pi\cos\theta$$
In terms of original coordinates, the integral we seek is
$$\frac{2\pi(z-vt)}{\sqrt{x^2 + y^2 + (z-vt)^2}}$$
