Howto evaluate this integral? $$
{1 \over \,\sqrt{\,2\pi^{3}\,}\,}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\mathrm{e}^{-\left(m^{2} + z^{2}\right)/2} \over
\,\sqrt{\,m^{2} + z^{2}\,}\,}\, mz\,\,\mathrm{d}m\,\mathrm{d}z
$$
I was told that this is an odd function, but I don't understand what that is/ how so. 
 A: Hint. One is allowed to applied Fubini's theorem here, obtaining
$$
\begin{align}
&\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} mz\cdot(2\pi^3)^{-1/2} \cdot(m^2+z^2)^{-1/2}\cdot e^{-(m^2+z^2)/2} dmdz
\\\\&=(2\pi^3)^{-1/2} \int_{-\infty}^{\infty} m\cdot e^{-m^2/2}\left(\int_{-\infty}^{\infty}\frac{z}{(m^2+z^2)^{1/2}}\cdot e^{-z^2/2} dz\right)dm
\\\\&=0
\end{align}
$$ due to 
$$
\int_{-\infty}^{\infty}\frac{z}{(m^2+z^2)^{1/2}}\cdot e^{-z^2/2} dz=0
$$ the latter integrand being an odd integrable function.
A: Switching to polar coordinates and applying Fubini we obtain that your integral equals:
$$\int_{-\pi}^{\pi}\cos(\phi)\sin(\phi)d\phi\cdot\frac1{\sqrt{2\pi^3}}\int_0^\infty r^2e^{-r^2/2}dr.$$
The second integral is convergent and the first is zero as the function is odd.
A: By a parity argument ($f(x,y)=-f(x,-y)$), the first/third and second/fourth quadrants exactly compensate each other.

A: Using a standard change of variable, the antiderivative  $$\int\frac{m\, z\, e^{-\frac{m^2+z^2}{2} }}{\sqrt{2} \pi ^{3/2}
   \sqrt{m^2+z^2}}\,dm=\frac{z }{2 \pi }\text{erf}\left(\frac{\sqrt{m^2+z^2}}{\sqrt{2}}\right)$$ which makes $$\int_{-\infty}^\infty \frac{m\, z\, e^{-\frac{m^2+z^2}{2} }}{\sqrt{2} \pi ^{3/2}
   \sqrt{m^2+z^2}}\,dm=0$$ which is unecessary to be done (refer to Yves Daoust's comments).
