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Consider the integral $$ I(x, q) = \int_0^{\text{arctanh}(x)} q^{-w} \text{tanh}(w) dw $$

Can this integral be simplified any further?

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The change of variable $z=\tanh(w)$ yields: $\displaystyle \color{red}{I(x,\mathrm e^{2r})=\int_0^x(1-z)^{r-1}(1+z)^{-r-1}z\,\mathrm dz} $.

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could you show how you got there? –  example Apr 9 '12 at 14:44
    
Ok, got it now. Had that idea as well, but did not see it through as far as you did oO Thank you =) –  example Apr 9 '12 at 15:25

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