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I stumbled into the following integral \begin{equation} \int_{\frac{c}{c+1}}^{\infty} \frac{1}{v^c(v+1)} dv. \hspace{2cm} \end{equation} Mathematica gives a solution for this $$ \int_{\frac{c}{c+1}}^{\infty} \frac{1}{v^c(v+1)} dv = (-1)^{-c}B\left(-\frac{1+c}{c};c,0 \right), $$ where $B(\cdot)$ is the incomplete beta function. But I can't figure out where this comes from. Indeed, the integral resembles the incomplete beta function, for which I found a property from Wikipedia: $B(x;a,b) = (-1)^aB(x/(x-1);a,1-a-b)$ – this somehow looks like Mathematica's solution, but I can't work it out.

For the indefinite integral Mathematica gives; $$ \int\frac{1}{v^c(v+1)} dv = \frac{v^{1-c} \, _2F_1(1,1-c;2-c;-v)}{c-1}, $$ where $_2F_1(\cdot)$ is the hypergeometric function. Does anyone know how this is derived?

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  • $\begingroup$ I found out that for $c=1$, we can make the transformation $v \mapsto 1/(1+v)$ for the beta function $B((1+c)/(1+2c),c,1-c)=\int_0^{(1+c)/(1+2c)} v^{c-1} (1-v)^{-c} dv,$ which is the given integral and the given Mathematica solution by the Wikipedia equation I gave. For general c, this seems to be complicated. $\endgroup$
    – Mundo
    Commented Apr 5, 2023 at 13:34

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Assume that $c>0$ and that $v^{c}$ means the principal value of $v^{c}$.

To get Mathematica's result directly, make the substitution $u = - \frac{1}{v}$.

Then $$ \begin{align} \int_{\frac{c}{1+c}}^{\infty} \frac{1}{v^c(v+1)} \, \mathrm dv &= \int_{-\frac{1+c}{c}}^{0} \left(- \frac{1}{u} \right)^{-c} \left(\frac{u-1}{u} \right)^{-1} \frac{\mathrm du}{u^{2}} \\ &= (-1)^{c} \int^{0}_{-\frac{1+c}{c}} u^{c-1} \left(u-1 \right)^{-1} \, \mathrm du \\ &=(-1)^{c} \int_{0}^{-\frac{1+c}{c}} u^{c-1} \left(1-u\right)^{-1} \, \mathrm du \\ &= (-1)^{c} \, B \left(- \frac{1+c}{c}; c, 0 \right), \end{align}$$ where $(-1)^{c} = e^{i \pi c}$.

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