A Feynman parametrization integration by parts

AI Thread Summary
The discussion focuses on transforming a complex integral expression using Feynman parametrization and integration by parts. The initial expression involves a product of terms with denominators raised to the third power, which is simplified through an identity that represents one of the terms as an integral. This process leads to a new form where the denominators are squared and involve absolute values, facilitating easier integration. The transformation highlights the utility of Feynman parametrization in simplifying integrals in quantum field theory contexts. Ultimately, the discussion illustrates a method for achieving a more manageable integral form.
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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)
 
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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
 
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