# Feynman parametrization

Feynman parametrization is a technique for evaluating loop integrals which arise from Feynman diagrams with one or more loops. However, it is sometimes useful in integration in areas of pure mathematics as well.

## Formulas

Richard Feynman observed that:
    ${\frac {1}{AB}}=\int _{0}^{1}{\frac {du}{\left[uA+(1-u)B\right]^{2}}}$
which is valid for any complex numbers A and B as long as 0 is not contained in the line segment connecting A and B. The formula helps to evaluate integrals like:
    ${\begin{aligned}\int {\frac {dp}{A(p)B(p)}}&=\int dp\int _{0}^{1}{\frac {du}{\left[uA(p)+(1-u)B(p)\right]^{2}}}\\&=\int _{0}^{1}du\int {\frac {dp}{\left[uA(p)+(1-u)B(p)\right]^{2}}}.\end{aligned}}$
If A(p) and B(p) are linear functions of p, then the last integral can be evaluated using substitution.
More generally, using the Dirac delta function $\delta$:
    ${\begin{aligned}{\frac {1}{A_{1}\cdots A_{n}}}&=(n-1)!\int _{0}^{1}du_{1}\cdots \int _{0}^{1}du_{n}{\frac {\delta (1-\sum _{k=1}^{n}u_{k})\;}{\left(\sum _{k=1}^{n}u_{k}A_{k}\right)^{n}}}\\&=(n-1)!\int _{0}^{1}du_{1}\int _{0}^{u_{1}}du_{2}\cdots \int _{0}^{u_{n-2}}du_{n-1}{\frac {1}{\left[A_{1}u_{n-1}+A_{2}(u_{n-2}-u_{n-1})+\dots +A_{n}(1-u_{1})\right]^{n}}}.\end{aligned}}$
This formula is valid for any complex numbers A$_1$,...,A$_n$ as long as 0 is not contained in their convex hull.
Even more generally, provided that ${\text{Re}}(\alpha _{j})>0$ for all $1\leq j\leq n$:
    ${\frac {1}{A_{1}^{\alpha _{1}}\cdots A_{n}^{\alpha _{n}}}}={\frac {\Gamma (\alpha _{1}+\dots +\alpha _{n})}{\Gamma (\alpha _{1})\cdots \Gamma (\alpha _{n})}}\int _{0}^{1}du_{1}\cdots \int _{0}^{1}du_{n}{\frac {\delta (1-\sum _{k=1}^{n}u_{k})\;u_{1}^{\alpha _{1}-1}\cdots u_{n}^{\alpha _{n}-1}}{\left(\sum _{k=1}^{n}u_{k}A_{k}\right)^{\sum _{k=1}^{n}\alpha _{k}}}}$
where the Gamma function $\Gamma$ was used.

## Derivation

    ${\frac {1}{AB}}={\frac {1}{A-B}}\left({\frac {1}{B}}-{\frac {1}{A}}\right)={\frac {1}{A-B}}\int _{B}^{A}{\frac {dz}{z^{2}}}.$
By using the substitution $u=(z-B)/(A-B)$, we have $du=dz/(A-B)$, and $z=uA+(1-u)B$, from which we get the desired result
    ${\frac {1}{AB}}=\int _{0}^{1}{\frac {du}{\left[uA+(1-u)B\right]^{2}}}.$
In more general cases, derivations can be done very efficiently using the Schwinger parametrization. For example, in order to derive the Feynman parametrized form of ${\frac {1}{A_{1}...A_{n}}}$, we first reexpress all the factors in the denominator in their Schwinger parametrized form:
    ${\frac {1}{A_{i}}}=\int _{0}^{\infty }ds_{i}\,e^{-s_{i}A_{i}}\ \ {\text{for }}i=1,\ldots ,n$
and rewrite,
    ${\frac {1}{A_{1}\cdots A_{n}}}=\int _{0}^{\infty }ds_{1}\cdots \int _{0}^{\infty }ds_{n}\exp \left(-\left(s_{1}A_{1}+\cdots +s_{n}A_{n}\right)\right).$
Then we perform the following change of integration variables,
    $\alpha =s_{1}+...+s_{n},$
    $\alpha _{i}={\frac {s_{i}}{s_{1}+\cdots +s_{n}}};\ i=1,\ldots ,n-1,$
to obtain,
    ${\frac {1}{A_{1}\cdots A_{n}}}=\int _{0}^{1}d\alpha _{1}\cdots d\alpha _{n-1}\int _{0}^{\infty }d\alpha \ \alpha ^{n-1}\exp \left(-\alpha \left\{\alpha _{1}A_{1}+\cdots +\alpha _{n-1}A_{n-1}+\left(1-\alpha _{1}-\cdots -\alpha _{n-1}\right)A_{n}\right\}\right).$
where ∫$_0$$^1$dα$_1$⋯dα$_{n−1}${\textstyle \int _{0}^{1}d\alpha _{1}\cdots d\alpha _{n-1}} denotes integration over the region $0\leq \alpha _{i}\leq 1$ with ∑$_{i=1}$$^{n−1}$α$_i$≤1{\textstyle \sum _{i=1}^{n-1}\alpha _{i}\leq 1}.
The next step is to perform the $\alpha$ integration.
    $\int _{0}^{\infty }d\alpha \ \alpha ^{n-1}\exp(-\alpha x)={\frac {\partial ^{n-1}}{\partial (-x)^{n-1}}}\left(\int _{0}^{\infty }d\alpha \exp(-\alpha x)\right)={\frac {\left(n-1\right)!}{x^{n}}}.$
where we have defined $x=\alpha _{1}A_{1}+\cdots +\alpha _{n-1}A_{n-1}+\left(1-\alpha _{1}-\cdots -\alpha _{n-1}\right)A_{n}.$
Substituting this result, we get to the penultimate form,
    ${\frac {1}{A_{1}\cdots A_{n}}}=\left(n-1\right)!\int _{0}^{1}d\alpha _{1}\cdots d\alpha _{n-1}{\frac {1}{[\alpha _{1}A_{1}+\cdots +\alpha _{n-1}A_{n-1}+\left(1-\alpha _{1}-\cdots -\alpha _{n-1}\right)A_{n}]^{n}}},$
and, after introducing an extra integral, we arrive at the final form of the Feynman parametrization, namely,
    ${\frac {1}{A_{1}\cdots A_{n}}}=\left(n-1\right)!\int _{0}^{1}d\alpha _{1}\cdots \int _{0}^{1}d\alpha _{n}{\frac {\delta \left(1-\alpha _{1}-\cdots -\alpha _{n}\right)}{[\alpha _{1}A_{1}+\cdots +\alpha _{n}A_{n}]^{n}}}.$
Similarly, in order to derive the Feynman parametrization form of the most general case, 1/A$_1$$^{α1}$...A$_n$$^{αn}$ {\textstyle {\frac {1}{A_{1}^{\alpha _{1}}...A_{n}^{\alpha _{n}}}}} one could begin with the suitable different Schwinger parametrization form of factors in the denominator, namely,
    ${\frac {1}{A_{1}^{\alpha _{1}}}}={\frac {1}{\left(\alpha _{1}-1\right)!}}\int _{0}^{\infty }ds_{1}\,s_{1}^{\alpha _{1}-1}e^{-s_{1}A_{1}}={\frac {1}{\Gamma (\alpha _{1})}}{\frac {\partial ^{\alpha _{1}-1}}{\partial (-A_{1})^{\alpha _{1}-1}}}\left(\int _{0}^{\infty }ds_{1}e^{-s_{1}A_{1}}\right)$
and then proceed exactly along the lines of previous case.

## Alternative form

An alternative form of the parametrization that is sometimes useful is
    ${\frac {1}{AB}}=\int _{0}^{\infty }{\frac {d\lambda }{\left[\lambda A+B\right]^{2}}}.$
This form can be derived using the change of variables $\lambda =u/(1-u)$. We can use the product rule to show that $d\lambda =du/(1-u)^{2}$, then
    ${\begin{aligned}{\frac {1}{AB}}&=\int _{0}^{1}{\frac {du}{\left[uA+(1-u)B\right]^{2}}}\\&=\int _{0}^{1}{\frac {du}{(1-u)^{2}}}{\frac {1}{\left[{\frac {u}{1-u}}A+B\right]^{2}}}\\&=\int _{0}^{\infty }{\frac {d\lambda }{\left[\lambda A+B\right]^{2}}}\\\end{aligned}}$
More generally we have
    ${\frac {1}{A^{m}B^{n}}}={\frac {\Gamma (m+n)}{\Gamma (m)\Gamma (n)}}\int _{0}^{\infty }{\frac {\lambda ^{m-1}d\lambda }{\left[\lambda A+B\right]^{n+m}}},$
where $\Gamma$ is the gamma function.
This form can be useful when combining a linear denominator $A$ with a quadratic denominator $B$, such as in heavy quark effective theory (HQET).

## Symmetric form

A symmetric form of the parametrization is occasionally used, where the integral is instead performed on the interval $[-1,1]$, leading to:
    ${\frac {1}{AB}}=2\int _{-1}^{1}{\frac {du}{\left[(1+u)A+(1-u)B\right]^{2}}}.$