# How to verify integral with hypergeometric function

Trying to evaluate the following integral, Mathematica returns this result:

$$\int \frac{e^{-\tau \omega}}{1+e^{-\beta \omega}} d \omega = \frac{e^{(\beta - \tau) \omega} \cdot {}_2F_1(1, 1-\frac{\tau}{\beta}, 2 - \frac{\tau}{\beta}, -e^{\beta \omega})}{\beta - \tau}$$

$\beta$ and $\tau$ can be treated as constants at this point. Unfortunately, I do not have any clue how I could achieve the same result with pen and paper. Does anyone have an idea?

-

Expand rhs into series and differentiate. Or to make a change of variable in lhs $t=e^{-\beta \omega}$, expand into series and integrate.

-
Thanks for the answer, I tried your second variant before asking here, but changing the variable does not work because the exponents are different and doing the full power series expansion leaves me with a mess of indices that ignore every attempt of mine to reorder them in powers of omega to make a term-wise integration possible. –  Alex Brentler Jun 19 '11 at 16:24
It is not needed to expand the exponent: $$\frac{e^{-\tau\omega}}{1+e^{-\beta\omega}}\,d\omega=-\frac{t^{1-\tau/\beta}}{1‌​+t}\,dt.$$ –  Andrew Jun 19 '11 at 20:39
Sorry, $$\frac{e^{-\tau\omega}}{1+e^{-\beta\omega}}=\frac{t^{\tau/\beta-1}}{1+t}\,dt.$$ –  Andrew Jun 19 '11 at 20:49
Thanks a lot, after doing the transformation to $t=e^{(\beta-\tau) \omega}$, $e^{\beta \omega} = t^{\frac{1}{1-\frac{\tau}{\beta}}}$ and remembering that $$d\omega = \frac{dt}{(\beta-\tau) t}$$ I could evaluate the integral using a series expansion. –  Alex Brentler Jun 21 '11 at 9:55
Differentiate the RHS to get the integrand on the LHS? Note: $$\frac{d}{dz} {}_2F_1(a,b;c;z) = \frac{ab}{c}\;{}_2F_1(a+1,b+1;c+1;z)$$