Expansion of nonregularized integral

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SUMMARY

The discussion focuses on the expansion of the nonregularized integral I_{\mu\nu}=\int d^{4}l\frac{4l_{\mu}l_{\nu}-g_{\mu\nu}l^2}{(l^{2}-B+i\epsilon)^{3}} into two distinct terms. The first term is expressed as 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}, while the second term involves the metric tensor g_{\mu\nu} and the integral of B. The calculation of the first term is aimed to yield the result -g_{\mu\nu}i\pi^{2}/2, utilizing tensor calculus and the chain rule.

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baranas
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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.
 
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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]
 

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