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.

Richard Feynman observed that:

$\frac{1}{AB}=\int^1_0 \frac{du}{\left[uA +(1-u)B\right]^2}$

which simplifies evaluating integrals like:

$\int \frac{dp}{A(p)B(p)}=\int dp \int^1_0 \frac{du}{\left[uA(p)+(1-u)B(p)\right]^2}=\int^1_0 du \int \frac{dp}{\left[uA(p)+(1-u)B(p)\right]^2}.$

More generally, using the Dirac delta function:

$\begin{align} \frac{1}{A_1\cdots A_n}&=(n-1)!\int^1_0 du_1 \cdots \int^1_0 du_n \frac{\delta(u_1+\dots+u_n-1)}{\left[u_1 A_1+\dots +u_n A_n\right]^n} \\ &=(n-1)! \int^1_0 du_1 \int^{u_1}_0 du_2 \cdots \int^{u_{n-2}}_0 du_{n-1} \frac{1}{\left[A_1+u_1(A_2-A_1)+\dots+u_{n-1} (A_n-A_{n-1})\right]^n}. \end{align}$

Even more generally, provided that $\text{Re} ( \alpha_{j} ) > 0$ for all $1 \leq j \leq n$:

$\frac{1}{A_{1}^{\alpha_{1}}\cdots A_{n}^{\alpha_{n}}}=\frac{\Gamma(\alpha_{1}+\dots+\alpha_{n})}{\Gamma(\alpha_{1})\cdots\Gamma(\alpha_{n})}\int_{0}^{1}du_{1}\cdots\int_{0}^{1}du_{n}\frac{\delta(\sum_{k=1}^{n}u_{k}-1)u_{1}^{\alpha_{1}-1}\cdots u_{n}^{\alpha_{n}-1}}{\left[u_{1}A_{1}+\cdots+u_{n}A_{n}\right]^{\sum_{k=1}^{n}\alpha_{k}}} .$

See also Schwinger parametrization.

Derivation

$\frac{1}{AB} = \frac{1}{A-B}\left(\frac{1}{B}-\frac{1}{A}\right)=\frac{1}{A-B}\int_B^A \frac{dz}{z^2}.$

Now just linearly transform the integral using the substitution,

$u=(z-B)/(A-B)$ which leads to $du = dz/(A-B)$ so $z = uA + (1-u)B$

and 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:

$\frac{1}{A_i}= \int^\infty_0 ds_i \, e^{-s_i A_i}, \forall i =1,...,n$

and rewrite,

$\frac{1}{A_1...A_n}=\int_0^\infty ds_1...ds_n \exp\left[-\left(s_1A_1+...s_nA_n\right)\right].$

Then we perform the following change of integration variables,

$\alpha = s_1+...+s_n,$

$\alpha_{i} = \frac{s_{i}}{s_1+...s_n}; i=1,\ldots,n-1,$

to obtain,

$\frac{1}{A_1...A_n} = \int_{0}^{1}d\alpha_1...d\alpha_{n-1} \int_{0}^{\infty}d\alpha \alpha^{N-1}\exp\left(-\alpha\left\{ \alpha_1A_1+...\alpha_{n-1}A_{n-1}+ \left(1-\alpha_{1}-...\alpha_{n-1}\right)A_{n}\right\} \right).$

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 $x= \alpha_1A_1+...\alpha_{n-1}A_{n-1}+ \left(1-\alpha_{1}-...\alpha_{n-1}\right)A_{n}.$

Substituting this result, we get to the penultimate form,

$\frac{1}{A_1...A_n}=\left(n-1\right)!\int_{0}^{1}d\alpha_1...d\alpha_{n-1}\frac{1}{[\alpha_1A_1+...+\alpha_{n-1}A_{n-1}+ \left(1-\alpha_{1}-...\alpha_{n-1}\right)A_{n}]^n} ,$

and, after introducing an extra integral, we arrive at the final form of the Feynman parametrization, namely,

$\frac{1}{A_1...A_n}=\left(n-1\right)!\int_{0}^{1}d\alpha_1...d\alpha_{n}\frac{\delta\left(1-\alpha_1-...-\alpha_n\right)}{[\alpha_1A_1+...+\alpha_{n}A_{n}]^n} .$

Similarly, in order to derive the Feynman parametrization form of the most general case, :$\frac{1}{A_1^{\alpha_1}...A_n^{\alpha_n}}$ one could begin with the suitable different Schwinger parametrization form of factors in the denominator, namely,

$\frac{1}{A_1^{\alpha_1}} = \frac{1}{\left(\alpha_1-1\right)!}\int^\infty_0 ds_1 \,s_1^{\alpha_1-1} e^{-s_1 A_1} = \frac{1}{\Gamma(\alpha_1)}\frac{\partial^{\alpha_1-1}}{\partial(-A_1)^{\alpha_1-1}}\left(\int_{0}^{\infty}ds_1 e^{-s_1 A_1}\right)$

and then proceed exactly along the lines of previous case.

Alternative Form

An alternative form of the parametrisation 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

$\begin{align} \frac{1}{AB} & = \int^1_0 \frac{du}{\left[uA +(1-u)B\right]^2} \\ & = \int^1_0 \frac{du}{(1-u)^{2}} \frac{1}{\left[\frac{u}{1-u} A + B \right]^2} \\ & = \int_{0}^{\infty} \frac{d\lambda}{\left[\lambda A + B\right]^2} \\ \end{align}$

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 from 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:

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