Question 4.7: Feynman Parametrization Prove the most general form of the F......
Feynman Parametrization
Prove the most general form of the Feynman parametrization,
\prod\limits_{i=1}^{n} \frac{1}{a_{i}^{A_{i}}}=\frac{\Gamma(A)}{\prod_{i=1}^{n} \Gamma\left(A_{i}\right)} \int_{0}^{1} \frac{\prod_{i=1}^{n} \mathrm{~d} x_{i} x_{i}^{A_{i}-1}}{\left(\sum_{i=1}^{n} a_{i} x_{i}\right)^{A}} \delta\left(1-\sum\limits_{i=1}^{n} x_{i}\right) (1)
with A=\sum_{i=1}^{n} A_{i} by means of mathematical induction. The a_{i}(i=1,2, \ldots, n) are arbitrary complex numbers.
To prove (1) we start with the simple formula
\frac{1}{a \cdot b}=\int\limits_{0}^{1} \frac{\mathrm{d} x}{(a x-b(1-x))^{2}}, (2)
which is obtained by observing that
\frac{1}{a \cdot b}=\frac{1}{b-a}\left(\frac{1}{a}-\frac{1}{b}\right)=\frac{1}{b-a} \int\limits_{a}^{b} \frac{\mathrm{d} z}{z^{2}}. (3)
Substituting z=a x+b(1-x) one obtains (2). Equation (2) can be further generalized to arbitrary powers by taking derivatives
\frac{\partial^{A}}{\partial a^{A}} \frac{\partial^{B}}{\partial b^{B}} (4)
on both sides so that
\begin{aligned} \frac{1}{a^{A} \cdot b^{B}} & =\frac{\Gamma(A+B)}{\Gamma(A) \Gamma(B)} \int\limits_{0}^{1} \mathrm{~d} x \frac{x^{A-1}(a-x)^{B-1}}{(a x+b(1-x))^{A+B}} \\ & =\frac{\Gamma(A+B)}{\Gamma(A) \Gamma(B)} \int\limits_{0}^{1} \mathrm{~d} x \mathrm{~d} y \frac{x^{A-1} y^{B-1}}{(a x+b y)^{A+B}} \delta(1-x-y) . & (5) \end{aligned}
Thus we have proven (1) for n=2. Now we assume that the formula (1) holds for n and prove that it is also valid for n+1. Let us start with the expression
\frac{1}{a_{n+1}^{A_{n+1}}} \prod_{i=1}^{n} \frac{1}{a_{i}^{A_{i}}}=\frac{\Gamma(A)}{\prod_{i=1}^{n} \Gamma\left(A_{i}\right)} \int_{0}^{1} \frac{\prod_{i=1}^{n} \mathrm{~d} x_{i} x_{i}^{A_{i}-1}}{\left(\sum_{i=1}^{n} a_{i} x_{i}\right)^{A}} \delta\left(1-\sum\limits_{i=1}^{n} x_{i}\right) \frac{1}{a_{n+1}^{A_{n+1}}}, (6)
where we multiplied (1) on both sides by 1 / a_{n+1}^{A_{n+1}}. Applying (5) we can rewrite (6) to obtain
\begin{aligned} \prod_{i=1}^{n} \frac{1}{a_{i}^{A_{i}}} & =\frac{\Gamma(A)}{\prod_{i=1}^{n} \Gamma\left(A_{i}\right)} \int\limits_{0}^{1} \prod_{i=1}^{n} \mathrm{~d} x_{i} x_{i}^{A_{i}-1} \delta\left(1-\sum\limits_{i=1}^{n} x_{i}\right) \frac{\Gamma\left(A+A_{n+1}\right)}{\Gamma(A) \Gamma\left(A_{n+1}\right)} \\ & \times \int\limits_{0}^{1} \mathrm{~d} x_{n+1} \mathrm{~d} y \frac{y^{A-1} x_{n+1}^{A_{n+1}-1}}{\left(y \sum_{i=1}^{n} a_{i} x_{i}+a_{n+1} x_{n+1}\right)^{A+A_{n+1}}} \delta\left(1-y-x_{n+1}\right) . & (7) \end{aligned}
Now we change variables \tilde{x}_{i}=y x_{i}(i=1,2, \ldots n) in the above equation and perform the y integration. After replacing \tilde{x}_{i} by x_{i} we obtain
\prod\limits_{i=1}^{n+1} \frac{1}{a_{i}^{A_{i}}}=\frac{\Gamma\left(A+A_{n+1}\right)}{\prod_{i=1}^{n+1} \Gamma\left(A_{i}\right)} \int\limits_{0}^{1} \frac{\prod_{i=1}^{n+1} \mathrm{~d} x_{i} x_{i}^{A_{i}-1}}{\left(\sum_{i=1}^{n+1} a_{i} x_{i}\right)^{A+A_{n+1}}} \delta\left(1-\sum\limits_{i=1}^{n+1} x_{i}\right) (8)
This completes the proof of (1).