# Expansion of nonregularized integral

1. May 19, 2012

### baranas

Can anyone help me with the expansion of the integral
$$I_{\mu\nu}=\int d^{4}l\frac{4l_{\mu}l_{\nu}-g_{\mu\nu}l^2}{(l^{2}-B+i\epsilon)^{3}}$$.
I would like to know how it could be expanded into two terms
$$I_{\mu\nu}=\frac{1}{2}\int d^{4}l\frac{\partial^2}{\partial l^\mu \partial l^\nu}\frac{1}{l^{2}-B+i\epsilon}-g_{\mu\nu}\int d^{4}l\frac{B}{(l^{2}-B+i\epsilon)^{3}}$$-.
Another question is how can i calculate the first term to have result
$$-g_{\mu\nu}i\pi^{2}/2$$.
P.S. I am inexperienced with tensor calculations. I would be grateful for any help.

Last edited: May 19, 2012
2. May 20, 2012

### baranas

It is all just a chain rule :)
$$\frac{1}{2}\partial_{\mu}\partial_{\nu}\frac{1}{l^{2}-\triangle+i\epsilon}=\partial_{\mu}\left[\frac{-l_{\nu}}{\left[l^{2}-\triangle+i\epsilon\right]^{2}}\right]$$
$$=\left[-g_{\mu\nu}\frac{l^{2}-\triangle}{\left[l^{2}-\triangle+i\epsilon\right]^{3}}+\frac{4l_{\nu}l_{\mu}}{\left[l^{2}-\triangle+i\epsilon\right]^{3}}\right]$$