A Triple integral While trying to obtain a Green's function for a PDE,I stumbled upon this integral
$\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z}$
How can I evaluate this triple integral?
 A: You can use the Inverse Fourier Transform to evaluate the 3 integrals.  In the derivation below, I have only used the properties and transform pairs listed on this page: https://en.wikipedia.org/wiki/Fourier_transform
$$\begin{align*}&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{i(s_{x}x+s_{y}y+s_{z}z)}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{x}ds_{y}ds_{z}  \\
\\
&= \int_{-\infty}^{\infty} e^{i s_{x}x} \int_{-\infty}^{\infty} e^{i s_{y}y} \int_{-\infty}^{\infty}\frac{e^{is_{z}z}}{s_{x}^{2}+s_{y}^{2}+2as_{z}-2a^{2}}ds_{z}ds_{y}ds_{x} \\
\\
&=  \int_{-\infty}^{\infty} e^{i s_{x}x} \int_{-\infty}^{\infty} e^{i s_{y}y} \space 2\pi\dfrac{i}{2a}\dfrac{1}{2}\space\mathscr{F}^{-1}\left\{\frac{2}{i\left(s_{z} -\left[a - \frac{s_{x}^{2}+s_{y}^{2}}{2a}\right] \right)}\right\}ds_{y}ds_{x}\\
\\
&= \dfrac{i\pi}{2a} \int_{-\infty}^{\infty} e^{i s_{x}x} \int_{-\infty}^{\infty} e^{i s_{y}y} \space \mathrm{sgn}(z) \space e^{iz\left(a - \frac{s_{x}^{2}+s_{y}^{2}}{2a}\right) } ds_{y}ds_{x}\\
\\
&= \dfrac{i\pi}{2a} \mathrm{sgn}(z) \space e^{iaz} \int_{-\infty}^{\infty} e^{-i\frac{z}{2a}s_{x}^{2}} \space e^{i s_{x}x} \int_{-\infty}^{\infty} e^{-i\frac{z}{2a}s_{y}^{2}} \space e^{i s_{y}y} \space  ds_{y}ds_{x}\\
\\
&= \dfrac{i\pi}{2a} \mathrm{sgn}(z) \space e^{iaz} \int_{-\infty}^{\infty} e^{-i\frac{z}{2a}s_{x}^{2}} \space e^{i s_{x}x} \int_{-\infty}^{\infty} 2\pi e^{-i\frac{\pi}{4}}\sqrt{\frac{a/2z}{\pi}}\sqrt{\frac{\pi}{a/2z}}e^{-i\left(\frac{s_{y}^{2}}{4(a/2z)}-\frac{\pi}{4}\right)} \space \frac{1}{2\pi}e^{i s_{y}y} \space  ds_{y}ds_{x}\\
\\
&= \dfrac{i\pi}{2a} \mathrm{sgn}(z) \space e^{iaz} \int_{-\infty}^{\infty} e^{-i\frac{z}{2a}s_{x}^{2}} \space e^{i s_{x}x}\space 2\pi e^{-i\frac{\pi}{4}}\sqrt{\frac{a/2z}{\pi}}\mathscr{F}^{-1}\left\{\sqrt{\frac{\pi}{a/2z}}e^{-i\left(\frac{s_{y}^{2}}{4(a/2z)}-\frac{\pi}{4}\right)} \right\}ds_{x}\\
\\
&= \dfrac{i\pi}{2a} \mathrm{sgn}(z) \space e^{iaz} \int_{-\infty}^{\infty} e^{-i\frac{z}{2a}s_{x}^{2}} \space e^{i s_{x}x}\space 2\pi e^{-i\frac{\pi}{4}}\sqrt{\frac{a}{2\pi z}} e^{i\frac{a}{2z}y^2}ds_{x}\\
\\
&= \dfrac{i\pi}{2a} \mathrm{sgn}(z) \space e^{iaz} \space 2\pi e^{-i\frac{\pi}{4}}\sqrt{\frac{a}{2\pi z}} e^{i\frac{a}{2z}y^2} \int_{-\infty}^{\infty} e^{-i\frac{z}{2a}s_{x}^{2}} \space e^{i s_{x}x}\space ds_{x}\\
\\
&= \dfrac{i\pi}{2a} \mathrm{sgn}(z) \space e^{iaz} \space 2\pi e^{-i\frac{\pi}{4}}\sqrt{\frac{a}{2\pi z}} e^{i\frac{a}{2z}y^2} \space 2\pi e^{-i\frac{\pi}{4}}\sqrt{\frac{a}{2\pi z}} e^{i\frac{a}{2z}x^2} \\
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&= \pi^2\dfrac{\mathrm{sgn}(z)}{z} e^{iaz} \space e^{i\frac{a}{2z}\left(x^2+y^2\right)}\\
\\
\end{align*}$$
if I didn't make an error.
