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I have a problem with a step in a physics text book [1]. It claims that \begin{equation} 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) \end{equation} ($a,b \in \mathbb{R}$) is equal to, by a change of integration variables, \begin{equation} \Gamma(a+b-1)\int_0^1 d\lambda\, \lambda^{-a} (1-\lambda)^{-b} \end{equation} It seems clear that one should be using the integral representation of the Gamma function \begin{equation} \Gamma(z) = \int_0^{\infty} dx\, x^{z-1} e^{-x} \end{equation} 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.

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  • $\begingroup$ Look at the Beta function integral representation en.wikipedia.org/wiki/Beta_function $\endgroup$ Dec 13 '19 at 13:44
  • $\begingroup$ 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? $\endgroup$ Dec 13 '19 at 13:54
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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 $a+b>1$.

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