# Evaluating the integral $\int_{c/(1+c)}^{\infty}1/(v^c(v+1))dv$ ("generalized incomplete beta function")

I stumbled into the following integral $$$$\int_{\frac{c}{c+1}}^{\infty} \frac{1}{v^c(v+1)} dv. \hspace{2cm}$$$$ Mathematica gives a solution for this $$\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?

• 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. Commented Apr 5, 2023 at 13:34

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 $$(-1)^{c} = e^{i \pi c}$$.