Convolution of plane wave with retarded propagator I am trying to compute the integral
$$
I(t,\mathbf{x},\mathbf{p}) := \int_0^\infty dt' \int_{\mathbb{R}^3} d^3\mathbf{x}'\  G(t,t',\mathbf{x},\mathbf{x}') e^{- i |\mathbf{p}| t' + i \mathbf{p} \cdot \mathbf{x}'}
$$
where $G(t,t',\mathbf{x},\mathbf{x}') = - \frac{\Theta(t-t')}{4\pi |\mathbf{x} - \mathbf{x}'|} \delta(t-t'-|\mathbf{x}-\mathbf{x}'|)$ is the retarded Green's function for the wave equation.
I am able to do the $t'$-integral and get
$$
I(t,\mathbf{x},\mathbf{p}) = - \frac{1}{4\pi} e^{- i |\mathbf{p}| t} \int_{\mathbb{R}^3} d^3\mathbf{x}'\  \Theta(t - |\mathbf{x} - \mathbf{x}'|) \frac{e^{+i |\mathbf{p}| |\mathbf{x} - \mathbf{x}'|} e^{+i \mathbf{p} \cdot \mathbf{x}'}}{|\mathbf{x} - \mathbf{x}'|} \ .
$$
However I get stuck at this point.
How does one make progress on this integral? This is a convolution with a plane-wave, and so a very simple solution to the wave equation, so I am thinking this calculation must exist somewhere but I have not been able to track it down.
 A: $\newcommand{\bm}[1]{\boldsymbol{#1}}
\newcommand{\ee}{\mathrm{e}}
\newcommand{\ii}{\mathrm{i}}
\newcommand{\dd}{\mathrm{d}}$Note that the step function implies that the integration over $\bm{x}'$ is restricted to a ball of radius $t$ around $\bm{x}$. Therefore, we can write $\bm{x}' = \bm{x} + \bm{y}$ and introduce spherical coordinates for $\bm{y}$ (with the third axis along $\bm{p}$) to obtain
\begin{align}
I(t,\bm{x},\bm{p}) &= - \frac{\operatorname{\Theta}(t) \ee^{\ii (\bm{p} \bm{\cdot} \bm{x} - \lvert \bm{p}\rvert t)}}{4 \pi} \int \limits_{B_t(\bm{x})} \dd^3 \bm{x}' \, \frac{\ee^{\ii [\lvert \bm{p} \rvert \lvert \bm{x}' - \bm{x}\rvert + \bm{p} \bm{\cdot} (\bm{x}' - \bm{x})]}}{\lvert \bm{x}' - \bm{x}\rvert} = - \frac{\operatorname{\Theta}(t) \ee^{\ii (\bm{p} \bm{\cdot} \bm{x} - \lvert \bm{p}\rvert t)}}{4 \pi} \int \limits_{B_t(0)} \dd^3 \bm{y} \, \frac{\ee^{\ii (\lvert \bm{p} \rvert \lvert \bm{y}\rvert + \bm{p} \bm{\cdot} \bm{y})}}{\lvert \bm{y}\rvert} \\
&= - \frac{\operatorname{\Theta}(t) \ee^{\ii (\bm{p} \bm{\cdot} \bm{x} - \lvert \bm{p}\rvert t)}}{4 \pi} \int \limits_0^t \dd r \int \limits_{-1}^1 \dd u \int \limits_0^{2 \pi} \dd \phi \, r \, \ee^{\ii (1 + u)\lvert \bm{p} \rvert r} = \frac{\ii \operatorname{\Theta}(t) \ee^{\ii (\bm{p} \bm{\cdot} \bm{x} - \lvert \bm{p}\rvert t)}}{2 \lvert \bm{p} \rvert} \int \limits_0^t \dd r \, (\ee^{2 \ii \lvert \bm{p} \rvert r}-1) \\
&= \frac{\ii \ee^{\ii \bm{p} \bm{\cdot} \bm{x}}}{2 \lvert \bm{p} \rvert^2} \left[\sin(\lvert \bm{p} \rvert t) - \lvert \bm{p} \rvert t \ee^{-\ii \lvert \bm{p} \rvert t}\right]\operatorname{\Theta}(t) \, .
\end{align}
We can check the result by computing
$$ \left(\Delta_\bm{x} - \partial_t^2\right) I(t,\bm{x},\bm{p}) = \ee^{\ii(\bm{p} \bm{\cdot} \bm{x} - \lvert \bm{p} \rvert t)} $$
for $t > 0$.
