How to derive this integral identity? There is a particular integral which comes up frequently in physics which a friend and I spent hours trying to figure out with no success. I don't know if this integral has a name so googling has been fruitless. Arfken states the solution in the back flap but I have no idea how to find it
$$\int_{-\infty}^{\infty}{\frac{e^{i\mathbf{k}^{.}\mathbf{r}}}{k^{2}+m^{2}}\frac{dk^{3}}{(2\pi)^{3}}}=\frac{e^{-mr}}{4\pi r}$$
Limits go from -inf to inf for all k. Note that k and r are vectors. Presumably r on the right hand side is $\lvert\mathbf{r}\rvert$ (I could be wrong).
If someone could show how this is derived or provide a link to its derivation it would be greatly appreciated.
 A: Align the vector $\vec r$ with the polar axis of $k$- space.  Then we can write
$$\begin{align}
\int_{\mathbb{R}^3}  \frac{e^{i\vec k\cdot\vec r}}{k^2+m^2}\,dk^3&=2\pi \int_0^\infty \int_0^\pi \frac{e^{ikr\cos(\theta)}}{k^2+m^2}\,k^2\sin(\theta)\,d\theta\,dk\\\\
&=\frac{4\pi}{r} \int_0^\infty \frac{k\sin(kr)}{k^2+m^2}\,dk\\\\
&=\frac{2\pi}{r}\,\text{Im}\,\int_{-\infty}^\infty \frac{ke^{ikr}}{k^2+m^2}\,dk\\\\
&=\frac{2\pi^2 e^{-mr}}{r}
\end{align}$$
A: We have the Fourier transform $$\int_{\mathbb{R}^3}\frac{e^{-mr-i\mathbf{k}\cdot\mathbf{r}}}{4\pi r}d^3\mathbf{r}=\frac{1}{2}\int_0^\pi d\theta\sin\theta\int_0^\infty e^{-(m+ik\cos\theta)r}r dr=\frac{1}{2}\int_0^\pi\frac{\sin\theta d\theta}{(m+ik\cos\theta)^2}.$$That's$$\frac{1}{2ik}[(m+ik\cos\theta)^{-1}]_0^\pi=\frac{1}{2ik}\bigg(\frac{1}{m-ik}-\frac{1}{m+ik}\bigg)=\frac{1}{m^2+k^2}.$$The result then follows from the inversion theorem.
A: As Mark Viola provided very concise and elegant solution, let me take the other direction and provide a solution of the generalized problem.
Consider a general case: Let $d\geq 1$, $\alpha > 0$, $m > 0$, and $\mathbf{r} \in \mathbb{R}^d$. Then
\begin{align*}
I
= I(d, \alpha, m, \mathbf{r})
:= \int_{\mathbb{R}^d} \frac{\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot\mathbf{r}}}{(\lvert\mathbf{k}\rvert^2 + m^2)^{\alpha}} \, \mathrm{d}\mathbf{k}.
\end{align*}
Then $I$ exists in improper sense, and moreover it turns out that $I$ admits a closed form in terms of the modified Bessel function of the second kind. As a first step, we can normalize the integral by applying the substitution $\mathbf{k} \mapsto m\mathbf{k}$:
$$ I(d, \alpha, m, r) = m^{d-2\alpha} I(d, \alpha, 1, m\mathbf{r}). $$
So we may assume $m = 1$ without losing the generality. Then
\begin{align*}
I(d, \alpha, 1, \mathbf{r})
&= \frac{1}{\Gamma(\alpha)} \int_{\mathbb{R}^d} \left( \int_{0}^{\infty} t^{\alpha-1}\mathrm{e}^{-(\lvert\mathbf{k}\rvert^2+1)t} \, \mathrm{d}t \right) \mathrm{e}^{\mathrm{i}\mathbf{k}\cdot\mathbf{r}} \, \mathrm{d}\mathbf{k} \\
&= \frac{1}{\Gamma(\alpha)} \int_{0}^{\infty} t^{\alpha-1} \mathrm{e}^{-t} \left( \int_{\mathbb{R}^d} \mathrm{e}^{-t\lvert\mathbf{k}\rvert^2+\mathrm{i}\mathbf{k}\cdot\mathbf{r}} \, \mathrm{d}\mathbf{k} \right) \, \mathrm{d}t \\
&= \frac{1}{\Gamma(\alpha)} \int_{0}^{\infty} t^{\alpha-1} \mathrm{e}^{-t} \left( \left(\frac{\pi}{t} \right)^{d/2} \mathrm{e}^{-\frac{\lvert\mathbf{r}\rvert^2}{4t}} \right) \, \mathrm{d}t \\
&= \frac{\pi^{\frac{d}{2}}}{\Gamma(\alpha)} \int_{0}^{\infty} t^{\alpha-1-\frac{d}{2}} \mathrm{e}^{-t-\frac{\lvert\mathbf{r}\rvert^2}{4t}} \, \mathrm{d}t.
\end{align*}
In the first line, we utilized the formula $\int_{0}^{\infty} t^{s-1}\mathrm{e}^{-at} \, \mathrm{d}t = \frac{\Gamma(\alpha)}{a^s}$ for $a, s > 0$, and in the third line we utilized the 1-D formula $\int_{\mathbb{R}} \mathrm{e}^{-tk^2 + \mathrm{i}kr} \, \mathrm{d}k = \sqrt{\frac{\pi}{t}} \mathrm{e}^{-\frac{r^2}{4t}} $ for $t > 0$ by utilizing the the $d$-D integral $\int_{\mathbb{R}^d} \mathrm{e}^{-t\lvert\mathbf{k}\rvert^2+\mathrm{i}\mathbf{k}\cdot\mathbf{r}} \, \mathrm{d}\mathbf{k}$ factors into the product of 1-dim integrals. Finally, applying the substitution $t = \frac{\lvert\mathbf{r}\rvert}{2}\mathrm{e}^{s}$ gives
\begin{align*}
I(d, \alpha, 1, \mathbf{r})
&= \frac{2\pi^{\frac{d}{2}}}{\Gamma(\alpha)} \left(\frac{\lvert\mathbf{r}\rvert}{2}\right)^{\alpha-\frac{d}{2}} \int_{0}^{\infty} \cosh\left(\left(\alpha-\tfrac{d}{2}\right)s\right) \mathrm{e}^{-\lvert\mathbf{r}\rvert\cosh s} \, \mathrm{d}s \\
&= \frac{2\pi^{\frac{d}{2}}}{\Gamma(\alpha)} \left(\frac{\lvert\mathbf{r}\rvert}{2}\right)^{\alpha-\frac{d}{2}} K_{\alpha-\frac{d}{2}}(\lvert\mathbf{r}\rvert),
\end{align*}
where $K$ is the modified Bessel function of the second kind. Summarizing,

$$ \int_{\mathbb{R}^d} \frac{\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot\mathbf{r}}}{(\lvert\mathbf{k}\rvert^2 + m^2)^{\alpha}} \, \mathrm{d}\mathbf{k}
= \frac{2\pi^{\frac{d}{2}}}{\Gamma(\alpha)} \left(\frac{\lvert\mathbf{r}\rvert}{2m}\right)^{\alpha-\frac{d}{2}} K_{\alpha-\frac{d}{2}}(m\lvert\mathbf{r}\rvert) \tag{1} $$

Now your question corresponds to $\alpha=1$ and $d = 3$. Using the formula $K_{-\frac{1}{2}}(x) = \sqrt{\frac{\pi}{2x}} \mathrm{e}^{-x}$, we obtain
$$ \int_{\mathbb{R}^3} \frac{\mathrm{e}^{\mathrm{i}\mathbf{k}\cdot\mathbf{r}}}{\lvert\mathbf{k}\rvert^2 + m^2} \, \mathrm{d}\mathbf{k}
= \frac{2\pi^2}{\lvert\mathbf{r}\rvert} \mathrm{e}^{-m\lvert\mathbf{r}\rvert}. $$
Dividing both sides by $(2\pi)^3$ yields the desired equality.
