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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?

2 Answers 2

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

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    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.2011-06-21
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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) $