Graduate Feynman parametrization integration by parts

[asmae]

TL;DR

Feynman parametrization integration by parts

How can i move from this expression:
$$\frac{4}{\pi^{4}} \int dk \frac{1}{k^2} \frac{1}{(1+i(k-k_{f}))^3} \frac{1}{(1+i(k-k_{i}))^3}$$
to this one:
$$\frac{4}{\pi^{4}} \int dk \frac{1}{k^2} \frac{1}{(1+|k-k_{i}|^2)^2} \frac{1}{(1+|k-k_{f}|^2)^2}$$
using Feynman parametrization (Integration by parts)

 

[nbryan5]

A:We start with \begin{align*}\frac{4}{\pi^{4}} \int dk \frac{1}{k^2} \frac{1}{(1+i(k-k_{f}))^3} \frac{1}{(1+i(k-k_{i}))^3}\end{align*}and use the identity\begin{align*}\frac{1}{(1+i(k-k_f))^3} = \int_0^1 du \,(1-u)^2 \, e^{i (k-k_f) u}\end{align*}to obtain\begin{align*}&\frac{4}{\pi^{4}} \int_0^1 du \,(1-u)^2 \, \int dk \frac{e^{i (k-k_f) u}}{k^2} \frac{1}{(1+i(k-k_{i}))^3} \\&= \frac{4}{\pi^{4}} \int_0^1 du \,(1-u)^2 \, \int dk \frac{e^{i (k-k_f) u - i(k-k_i)}}{k^2(1+i(k-k_i))^2} \\&= \frac{4}{\pi^{4}} \int_0^1 du \,(1-u)^2 \, \int dk \frac{e^{i k(u-1) -i k_i (1+u)}}{k^2(1+i(k-k_i))^2} \\&= \frac{4}{\pi^{4}} \int_0^1 du \,(1-u)^2 \, \int dk \frac{e^{i k(u-1)}}{k^2(1+i(k-k_i + k_i(1+u)))^2} \\&= \frac{4}{\pi^{4}} \int_0^1 du \,(1-u