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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.

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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.

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  • $\begingroup$ Hello Mr.Olivier. Thank you very much for the hint, and sorry me for the late answer. I need just some more clarification. I continue your reasoning, and I find as below: $$ \begin{align} I &= \frac{2c^{2/\alpha}}{\alpha}\int_{0}^1(1-t)^{-2/\alpha-1}t^{2/\alpha-1}dt - \frac{2c^{2/\alpha}}{\alpha}\int_{0}^1 (1-t)^{-2/\alpha-1}t^{n+2/\alpha-1}dt \\\\&= \frac{2c^{2/\alpha}}{\alpha} \left( B( \frac{2}{\alpha} ; \frac{-2}{\alpha} ) - B( n+ \frac{2}{\alpha} ; \frac{-2}{\alpha} ) \right) \end{align} $$ $\endgroup$
    – adil
    Feb 18 '17 at 12:25
  • $\begingroup$ No problem. If you write this in terms of $\Gamma$, you should obtain the announced result. $\endgroup$ Feb 18 '17 at 12:27
  • $\begingroup$ $$ I = \frac{2c^{2/\alpha}}{\alpha} \left( \frac{\Gamma(\frac{2}{\alpha}).\Gamma(\frac{-2}{\alpha}) }{\Gamma(0)} - \frac{\Gamma(n+ \frac{2}{\alpha}).\Gamma(\frac{-2}{\alpha}) }{\Gamma(n)} \right) $$ So the first term go to 0, like $ \Gamma(0) $ go to infinity. and with $ - \frac{2}{\alpha}. \Gamma(\frac{-2}{\alpha})= \Gamma(1 - \frac{2}{\alpha}) $ , we got the result. Is that approach true ? because I see that for Beta function B(x,y), Re(x) and Re(y) should be positifs. and for Gamma function, is it defined at 0 ? Can we talk about $ \Gamma(0) $ ? $\endgroup$
    – adil
    Feb 18 '17 at 12:42
  • $\begingroup$ @adil I've edited my hint. Thanks. $\endgroup$ Feb 18 '17 at 12:48

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