Integral Proof Challenge Here is the challenge integral appearing on a physics book:
To prove the following equality
$$\iiint_{\mathbb{R}^3}\frac{1}{(2\pi)^3}\frac{e^{-i\vec{q}\cdot\vec{r}}}{|\vec{q}|^2+m^2}dq_xdq_ydq_z=\frac{e^{-m|\vec{r}|}}{4\pi|\vec{r}|}
$$
Here, $\vec{r}=x\vec{i}+y\vec{j}+z\vec{k}$, $m$ is non-negative.
I think this should be a Fourier transformation, specifically, I find that change the sign of $\vec{q}$ can keep the left-hand side unchanged. Hence, one can have the integrand being $\frac{1}{(2\pi)^3}\frac{e^{i\vec{q}\cdot\vec{r}}}{|\vec{q}|^2+m^2}$, which is now a typical inverse Fourier transform with the transforming function being $\frac{1}{|\vec{q}|^2+m^2}$. However, when I do the transformation part by part, i.e. do $q_x$ then $q_y$ then $q_z$, after transforming $q_x$, the integrand will have a square root term which will be hard to continue. Then I get stuck in this way. On the other hand, I try to express the whole integral in the spherical coordinate, notice, one can show that this triple integral is unchanged under any rotation (specifically, given a rotation matrix $R$, then $\vec{q}\cdot\vec{r}$ and $|\vec{q}|^2$ is unchanged under action of $R$, and notice $d^3q:=dq_xdq_ydq_z$ under the transformed frame $d^3q’=det\,R d^3q$ where $R\in SO_3\Rightarrow det\, R=1$, hence it is unchanged). One can choose direction of $\vec{r}$ to be alone $z$-axis. Then, if my calculation is correct (which seems not true), one need to calculate $\int_0^\pi d\varphi e^{-m|\vec{r}|\cos(\varphi)}$ which does not have a simple original function.
Can someone gives me some hint? It can be other solution method or the spherical one. For the spherical one, could you point out which part is easy to make mistakes. Thanks in advance.  
 A: Caution. Within the purview of mathematics, the integral is not absolutely convergent and thus an appropriate interpretation must be introduced. In this case, the integral can be considered as the limit
$$ \lim_{R\to\infty} \frac{1}{(2\pi)^3} \int_{B_R(0)} \frac{e^{-i\langle q,r\rangle}}{\|q\|^2+m^2} \, \mathrm{d}q. $$
Now let me convince you that spherical coordinates works here. By the rotational symmetry we may assume that $r = (0, 0, \|r\|)$. Then
\begin{align*}
\frac{1}{(2\pi)^3} \int_{B_R(0)} \frac{e^{-i\langle q,r\rangle}}{\|q\|^2+m^2} \, \mathrm{d}q
&= \frac{1}{(2\pi)^3} \int_{0}^{R} \int_{0}^{\pi} \frac{e^{-i\rho \|r\| \cos \phi}}{\rho^2 + m^2} \, 2\pi\rho^2 \sin\phi \, \mathrm{d}\phi\mathrm{d}\rho \\
&= \frac{1}{(2\pi)^2} \int_{0}^{R} \frac{\rho^2}{\rho^2 + m^2} \left[ \frac{e^{-i\rho \|r\| \cos \phi}}{i\rho\|r\|} \right]_{0}^{\pi} \,\mathrm{d}\rho \\
&= \frac{1}{2\pi^2\|r\|} \int_{0}^{R} \frac{\rho \sin(\rho \|r\|)}{\rho^2 + m^2} \,\mathrm{d}\rho \\
&\xrightarrow[R\to\infty]{} \frac{1}{2\pi^2\|r\|} \int_{0}^{\infty} \frac{\rho \sin(\rho \|r\|)}{\rho^2 + m^2} \,\mathrm{d}\rho.
\end{align*}
Now the last integral can be evaluated in various ways, including contour integral, and we are led to
$$ \lim_{R\to\infty} \frac{1}{(2\pi)^3} \int_{B_R(0)} \frac{e^{-i\langle q,r\rangle}}{\|q\|^2+m^2} \, \mathrm{d}q
= \frac{1}{2\pi^2\|r\|} \cdot \frac{\pi}{2} e^{-m\|r\|}
= \frac{e^{-m\|r\|}}{4\pi \|r\|}. $$

Alternatively, one may plug
$$ \frac{1}{\|q\|^2 +m^2} = \int_{0}^{\infty} e^{-(\|q\|^2+m^2)s} \, \mathrm{d}s$$
and interchange the order of integration to obtain
\begin{align*}
\frac{1}{(2\pi)^3} \int_{\mathbb{R}^3} \frac{e^{-i\langle q,r\rangle}}{\|q\|^2+m^2} \, \mathrm{d}q
&= \frac{1}{(2\pi)^3} \int_{0}^{\infty} e^{-m^2s} \prod_{k=1}^{3} \left( \int_{\mathbb{R}} e^{-s q_k^2 - iq_k r_k} \, \mathrm{d}q_k \right) \, \mathrm{d}s \\
&= \frac{1}{(2\pi)^3} \int_{0}^{\infty} e^{-m^2s} \prod_{k=1}^{3} \left( \sqrt{\frac{\pi}{s}} e^{-r_k^2/4s} \right) \, \mathrm{d}s \\
(s=\|r\|^2/4t^2) \qquad &= \frac{1}{2\pi^{3/2}\|r\|} \int_{0}^{\infty} e^{ - t^2-\frac{m^2\|r\|^2}{4t^2}} \, \mathrm{d}t.
\end{align*}
Again, one may want to introduce some modification of this argument to justify applying the Fubini's theorem to a non-absolutely convergent integral. In this case, Gaussian regularization is enough. Finally, the last integral can be computed by the Glasser's master theorem.
