can someone help me to prove that the following integral can be writting as follows:
$$ I= \int_{0}^\infty (1- \frac{1}{(1+c\cdot r^{-\alpha})^n}) 2r\cdot dr = c^{2/\alpha}\cdot\frac{\Gamma(1-\frac{2}{\alpha})\cdot\Gamma(n+\frac{2}{\alpha})}{\Gamma(n)}$$
Many thanks in advance.
Hint. By making the change of variable $$ t=\frac{1}{1+c\cdot r^{-\alpha}}, \quad r=c^{1/\alpha}\left(\frac1t-1 \right)^{-1/\alpha}, \quad dr=\frac{c^{1/\alpha}}{\alpha}\cdot t^{-2}\cdot\left(\frac1t-1 \right)^{-1/\alpha-1}dt, $$ one gets $$ \begin{align} &\int_{0}^\infty \left(1- \frac{1}{(1+c\cdot r^{-\alpha})^n}\right) 2r\cdot dr \\\\&=-\frac{2c^{2/\alpha}}{\alpha}\int_{0}^1 (1- t^n)(1-t)^{-2/\alpha-1}t^{2/\alpha-1}dt \\\\&=\frac{2c^{2/\alpha}}{\alpha}\sum_{k=0}^{n-1}\int_{0}^1 (1-t)^{-2/\alpha}t^{k+2/\alpha-1}dt \\\\&=\frac{2c^{2/\alpha}}{\alpha}\sum_{k=0}^{n-1}\frac{\Gamma(1-\frac{2}{\alpha})\cdot\Gamma(k+\frac{2}{\alpha})}{\Gamma(k+1)}. \end{align} $$ where we have used the Euler beta function.