I'm trying to compute the following integral:
$$ I(z_1,z_2,z_3)=\int_{\mathbb{R}^3}\frac{\mathrm{d}\nu_1\mathrm{d}\nu_2\mathrm{d}\nu_3}{(2\pi)^3}\frac{e^\left(iz_1\nu_1+iz_2\nu_2+iz_3\nu_3\right)}{(\nu_1-\nu_2)(\nu_1-\nu_3)(\nu_2-\nu_3)}$$
where $z_1, z_2, z_3$ are supposed to be real-valued.
It is possible to integrate first on $\nu_3$ by doing a simple element decomposition on $\dfrac{1}{(\nu_1-\nu_3)(\nu_2-\nu_3)}$ and use the identity
$$\int_{\mathbb{R}}\frac{\mathrm{d}\nu}{2\pi}\frac{e^{i\nu y}}{\nu}=i\int_{\mathbb{R}^+}\frac{\mathrm{d}\nu}{\pi} \frac{\sin(\nu y)}{\nu }=\frac{i}{2}\; \mathrm{sgn}(y)$$
and then we use the same technique to integrate over $\nu_2$ for instance.
I was wondering if there exists a more elegant way to solve this integral without using the machinery of simple element decomposition. Maybe an elegant change of variable ?
This integral is the Fourier transform of the inverse of the Vandermonde determinant, maybe there is something to do with that.
Thanks !