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 A₁,...,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 ${\textstyle \int _{0}^{1}d\alpha _{1}\cdots d\alpha _{n-1}}$ denotes integration over the region $0\leq \alpha _{i}\leq 1$ with ${\textstyle \sum _{i=1}^{n-1}\alpha _{i}\leq 1}$.

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,${\textstyle {\frac {1}{A_{1}^{\alpha _{1}}...A_{n}^{\alpha _{n}}}}}$ 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:

${\frac {1}{AB}}=2\int _{-1}^{1}{\frac {du}{\left[(1+u)A+(1-u)B\right]^{2}}}.$