# Expressing an integral in terms of gamma functions

1. Dec 23, 2015

### CAF123

I want to show that $$\int_0^{\infty} \frac{ds}{s-q^2} \frac{s^{-1-\epsilon}}{s-t \frac{z}{1-z}} = \Gamma(1-\epsilon) \Gamma(\epsilon) \frac{1}{t \frac{z}{1-z} - q^2} \left((-t)^{-1-\epsilon} \left(\frac{z}{1-z}\right)^{-1-\epsilon} -(-q^2)^{-1-\epsilon}\right)$$

I have many ideas on how to obtain the final answer but I haven't quite got there. I can rexpress the two denominator terms in terms of partial fractions to give $$\frac{1}{q^2 - t\frac{z}{1-z}}\int_0^{\infty} ds \,s^{-1-\epsilon} \left( \frac{1}{s-q^2} - \frac{1}{s - t\frac{z}{1-z}}\right)$$

Consider the first integral: $$\int_0^{\infty} ds \, \frac{s^{-1-\epsilon}}{s-q^2}.$$ If I perform integration by parts I get $$\int_0^{\infty} ds \, \frac{s^{-1-\epsilon}}{s-q^2} = s^{-1-\epsilon}\ln|s-q^2| |^{\infty}_0 + (1+\epsilon) \int_0^{\infty}ds\, s^{-2-\epsilon} \ln |s-q^2|$$ I then thought about using the fact that $$\ln w = \frac{d}{d\alpha} w^{\alpha}|_{\alpha=0}$$ to have everything in terms of rationals but the subsequent integral is no more tractable as far as I can tell. I also thought about using 'Schwinger parameters' but again not quite of the form required.

Then I went back to the starting integral and thought about feynman parametrisation https://en.wikipedia.org/wiki/Feynman_parametrization to rexpress the two denominators in terms of an integral but again, the resulting integral did not lead anywhere.

Any hints or suggestions would be great! Thanks

2. Dec 28, 2015

### Greg Bernhardt

I tried working backwards from the form $\Gamma(1-\epsilon)\Gamma(\epsilon)$.
One of them was straightforward, the other produced $\int \frac{y^{\epsilon-1}(1+y)}{1+y^2}dy$. Not sure whether that can be got into the form you need, or maybe I made a mistake somewhere. Looks close though.