Evalulate $\int_0^\infty \int_0^\infty \int_0^\infty e^{-(xy+yz+zx)}\ dx\ dy\ dz$ I am wondering how to evaluate this integral. Wolfram Alpha says it is $\frac{\pi^{3/2}}2$ but I have no idea how get there.
$$\\
\int_0^\infty \int_0^\infty \int_0^\infty e^{-(xy+yz+zx)}\ dx\ dy\ dz\\
$$
I guess it may have some connection with
$$\int_0^\infty \int_0^\infty \int_0^\infty e^{-(x^2+y^2+z^2)}\ dx\ dy\ dz=\left(\int_0^\infty e^{-x^2}\ dx\right)^3=\frac{\pi^{3/2}}8$$
and/or
$$\int_0^\infty \int_0^\infty \int_0^\infty e^{-(x+y+z)^2}\ dx\ dy\ dz=\dots=\frac{\sqrt{\pi}}8$$
but I am not sure. Thanks in advance.

Taking @ersh's suggestion, I did:
\begin{align}
&\int_0^\infty \int_0^\infty \int_0^\infty e^{-(xy+yz+zx)}\ dx\ dy\ dz\\
&=\int_0^\infty \int_0^\infty e^{-yz}\int_0^\infty e^{-(y+z)x}\ dx\ dy\ dz\\
&=\int_0^\infty \int_0^\infty \frac{e^{-yz}}{y+z}\ dy\ dz=\int_0^\infty \int_z^\infty \frac{e^{-(y-z)z}}{y}\ dy\ dz\\
&=\int_0^\infty e^{z^2}\int_z^\infty \frac{e^{-yz}}{y}\ dy\ dz\\
\end{align}
Now let me work on this integral:
\begin{align}
I(a)&=\int_z^\infty \frac{e^{-ayz}}{y}\ dy\\
\frac{dI(a)}{da}&=\int_z^\infty ze^{-ayz}\ dy=\left[-\frac{e^{-ayz}}a\right]_z^\infty=\frac{e^{-az^2}}{a}\\
\lim_{a\to0}I(a)&=-z^2\\
\therefore I(a)&=
\end{align}
Wait. I'm in the loop!
 A: A chain of mostly elementary manipulations:
\begin{align*}I &= \int_0^{\infty}\int_0^{\infty}\int_0^{\infty}e^{-(xy+yz+zx)}\,dx\,dy\,dz\\
&= \int_0^{\infty}\int_0^{\infty}\int_0^{\infty}e^{-yz}e^{-x(y+z)}\,dx\,dy\,dz\\
&= \int_0^{\infty}\int_0^{\infty}\frac{e^{-yz}}{y+z}\,dy\,dz\end{align*}
First, we evaluate the inner integral with elementary techniques. There's some clutter that amounts to constant multipliers, but it's just $e^{ax}$ at its core. Next, to handle $y$, we substitute $y=zu$, $dy=z\,du$:
\begin{align*}I &= \int_0^{\infty}\int_0^{\infty} \frac{e^{-z^2u}}{z(1+u)}\cdot z\,du\,dz\\
&= \int_0^{\infty}\int_0^{\infty} \frac{e^{-uz^2}}{1+u}\,dz\,du\\
&= \int_0^{\infty}\frac{\frac{\sqrt{\pi}}{2\sqrt{u}}}{1+u}\,du\end{align*}
After the substitution, the $u$ integral isn't something we want to deal with - so we switch the order and do the $z$ integral next. That's just a Gaussian - no elementary antiderivative, but we know that $\int_0^{\infty} e^{-az^2}\,dz=\frac{\sqrt{\pi}}{2\sqrt{a}}$. This leaves us with a $u$ integral that's algebraic; we substitute $v=\sqrt{u}$, $dv=\frac1{2\sqrt{u}}\,du$ to rationalize it:
$$I = \sqrt{\pi}\int_0^{\infty}\frac1{1+v^2}\,dv = \sqrt{\pi}\cdot\frac{\pi}{2}=\frac{\pi^{\frac32}}{2}$$
Done.
There's probably a way to connect this to a three-dimensional Gaussian, but I found it easier to break it down this way.
A: Here's another approach: By making substitution $$
(u,v,w) = (xy,yz,zx),
$$ we have$$\mathrm du \mathrm dv\mathrm dw = 2xyz \mathrm dx\mathrm dy\mathrm dz\implies \frac{\mathrm du \mathrm dv\mathrm dw}{2\sqrt{uvw}}=\mathrm dx\mathrm dy\mathrm dz.$$ Hence,
$$\begin{align*}
\int_0^\infty \int_0^\infty \int_0^\infty e^{-(xy+yz+zx)}\mathrm dx\mathrm dy\mathrm dz&=\frac12\int_0^\infty \int_0^\infty \int_0^\infty \frac{e^{-u-v-w}\mathrm du \mathrm dv\mathrm dw}{\sqrt{uvw}}\\&=\frac12\left( \int_0^\infty \frac{e^{-u}\mathrm du}{\sqrt{u}}\right)^3\\&=\frac12\left(2 \int_0^\infty e^{-v^2}\mathrm dv\right)^3\\
&=\frac{\pi^{\frac32}}{2}.
\end{align*}$$
