# Turning an integral into a Gamma Function

I have a problem with a step in a physics text book [1]. It claims that $$$$I(a,b) = \int_0^\infty dt\int_0^\infty du\, \frac{t^{a-1}u^{b-1} }{t+u}\exp \left(-\frac{tu }{t+u}\right)$$$$ ($$a,b \in \mathbb{R}$$) is equal to, by a change of integration variables, $$$$\Gamma(a+b-1)\int_0^1 d\lambda\, \lambda^{-a} (1-\lambda)^{-b}$$$$ It seems clear that one should be using the integral representation of the Gamma function $$$$\Gamma(z) = \int_0^{\infty} dx\, x^{z-1} e^{-x}$$$$ and so one of the new integration variables if $$x = tu/(u+t)$$. But I can't figure out how to make it work.

Any help would be much appreciated.

[1] Green, Schwarz & Witten, Superstring Theory, Volume 1, p. 387.

• Look at the Beta function integral representation en.wikipedia.org/wiki/Beta_function Dec 13 '19 at 13:44
• It is clear to me that the integral over $\lambda$ is just the Beta function, but how do you get the other part to be just a Gamma function? Dec 13 '19 at 13:54

## 1 Answer

The new variables are $$x = \frac{tu}{t+u} \in (0,\infty)$$ (as you suggested) and $$\lambda = \frac{u}{t+u} \in (0,1)$$. Then $$t = \frac{x}{\lambda}$$, $$u = \frac{x}{1-\lambda}$$ and the Jacobian determinant equals $$\frac{x}{\lambda^2 (1-\lambda)^2}$$, so \begin{align} I(a,b) &= \int \limits_0^\infty \int \limits_0^\infty \frac{t^{a-1} u^{b-1}}{t+u} \, \mathrm{e}^{- \frac{t u}{t+u}} \, \mathrm{d} t \, \mathrm{d} u = \int \limits_0^1 \int \limits_0^\infty \left(\frac{x}{\lambda}\right)^{a-2} \left(\frac{x}{1-\lambda}\right)^{b-2} x \, \mathrm{e}^{-x} \frac{x}{\lambda^2 (1-\lambda)^2} \, \mathrm{d} x \, \mathrm{d} \lambda \\ &= \int \limits_0^\infty x^{a+b-2} \, \mathrm{e}^{-x} \, \mathrm{d} x \int \limits_0^1 \lambda^{-a} (1-\lambda)^{-b} \, \mathrm{d} \lambda = \operatorname{\Gamma} (a+b-1) \operatorname{B} (1-a,1-b) \, . \end{align} The integral is convergent if and only if $$a<1$$, $$b<1$$ and $$a+b>1$$.