Monthly Archives: September 2018

Schwinger parametrization

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Hello everyone, I’m back today with an article about an integration technique I just recently discovered that’s called Schwinger parametrization. It actually originated in the calculation of loop integrals in Quantum Field Theory.

The method makes use of the following identity, which can be derived by a change of variable in the usual definition of the Gamma function.

\begin{aligned} \frac{1}{\beta^m(x)}=\frac{1}{\Gamma(m)}\int_0^\infty u^{m-1} e^{-u\beta(x)}\,du \end{aligned}

It’s a nice way to deal with integrals involving arbitrary powers of a function in the denominator. In this post and with this technique, I’ll show you that

\begin{aligned} I(n,p)=\int_0^\infty \left(\frac{1}{1+x^n}\right)^p\,dx=\frac{\pi}{\Gamma(p) n}\csc\left(\frac{\pi}{n}\right)\prod_{k=1}^{p-1} \left(p-k-\frac{1}{n}\right) \end{aligned}

Now this can be a pretty hard integral to compute, but Schwinger parametrization allows us to find its value.

First, let’s Schwinger parametrize it, letting \beta(x)=\frac{1}{1+x^n}

\begin{aligned} \int _0^\infty \left(\frac{1}{1+x^n}\right)^p\,dx &= \int _0^\infty \frac{1}{\left(1+x^n\right)^p}\,dx \\ &= \frac{1}{\Gamma(p)}\int_0^\infty \int_0^\infty u^{p-1} e^{-u(1+x^n)} \,du dx \\ &= \frac{1}{\Gamma(p)}\int_0^\infty \int_0^\infty u^{p-1} e^{-u-ux^n} \,du dx \\ &= \frac{1}{\Gamma(p)}\int_0^\infty \int_0^\infty u^{p-1} e^{-u}e^{-ux^n} \,du dx \end{aligned}

Upon switching the order of integration, we find that

\begin{aligned} \frac{1}{\Gamma(p)}\int_0^\infty \int_0^\infty u^{p-1} e^{-u}e^{-ux^n} \,dx du &= \frac{1}{\Gamma(p)}\int_0^\infty u^{p-1} e^{-u}\,du\int_0^\infty e^{-ux^n} \,dx \\ &= \frac{1}{\Gamma(p)}\int_0^\infty u^{p-1} e^{-u}\,du \int_0^\infty \frac{s^{1/n -1}}{n} e^{-us} \,ds \\ &= \frac{1}{\Gamma(p)}\int_0^\infty u^{p-1} e^{-u}\,du \frac{\Gamma(\frac{1}{n})}{u^{1/n}n} \\ &=\frac{\Gamma\left(1+\frac{1}{n}\right)}{\Gamma(p)}\int_0^\infty u^{p-1/n-1} e^{-u}\,du \end{aligned}

Where we simply used the substitution s=x^n. This last integral can also be expressed in terms of the Gamma function:

\begin{aligned} \frac{\Gamma\left(1+\frac{1}{n}\right)}{\Gamma(p)}\int_0^\infty u^{p-1/n-1} e^{-u}\,du &= \frac{\Gamma\left(1+\frac{1}{n}\right)\Gamma\left(p-\frac{1}{n}\right) }{\Gamma(p)} \end{aligned}

And we have our first result (that will be simplified):

\begin{aligned} \int_0^\infty \frac{1}{\left(1+x^n\right)^p}\,dx=\frac{\Gamma(\frac{1}{n}+1)\Gamma(p-\frac{1}{n})}{\Gamma (p)} \text{ for } pn>1 \end{aligned}

For fixed values of p, we do have nice results:

\begin{aligned} I(n,1) &= \int_0^\infty \frac{1}{1+x^n}\,dx=\frac{\pi}{n}\csc\left(\frac{\pi}{n}\right) \\ I(n,2) &= \int_0^\infty \frac{1}{\left(1+x^n\right)^2}\,dx=\frac{\pi}{n^2}(n-1)\csc\left(\frac{\pi}{n}\right) \\ I(n,3) &= \int_0^\infty \frac{1}{\left(1+x^n\right)^3}\,dx=\frac{\pi}{2n^3}(n-1)(2n-1)\csc\left(\frac{\pi}{n}\right) \end{aligned}

This result can be generalized for any integer values of p using some facts about the Gamma function:

\begin{aligned} \frac{\Gamma(\frac{1}{n}+1)\Gamma\left(p-\frac{1}{n}\right)}{\Gamma (p)} &= \frac{\Gamma\left(1+\frac{1}{n})\right)\Gamma\left(1-\frac{1}{n}\right)}{\Gamma(p)}\prod_{k=1}^{p-1} \left(p-k-\frac{1}{n}\right) \end{aligned}

And using the fact that

\begin{aligned} \Gamma\left(1+\frac{1}{n}\right)\Gamma\left(1-\frac{1}{n}\right)= \frac{\pi}{n}\csc\left(\frac{\pi}{n}\right) \end{aligned}

We find

\begin{aligned} \int_0^\infty \left(\frac{1}{1+x^n}\right)^p\,dx=\frac{\pi}{\Gamma(p) n}\csc\left(\frac{\pi}{n}\right)\prod_{k=1}^{p-1} \left(p-k-\frac{1}{n}\right) \end{aligned}

Using this equality, we can even find the L^p norm over [0,\infty) of the function f(x)=\frac{1}{1+x^n}:

\begin{aligned} ||f||_{L^p}&=\left(\int_0^\infty |f|^p\,dx\right)^{1/p} \\ &= \left(\int_0^\infty \left(\frac{1}{1+x^n}\right)^p \,dx\right)^{1/p} \\ &= \sqrt[p]{\frac{\pi}{\Gamma(p) n}\csc\left(\frac{\pi}{n}\right)\prod_{k=1}^{p-1} \left(p-k-\frac{1}{n}\right)} \end{aligned}