# Expressing an integral in terms of gamma functions

Gold Member
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

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haruspex