I want to solve the following integration $$\int_0^{\infty}[1-(\frac{1}{1+x})^M]x^{-\frac{2}{\alpha}-1}dx$$ where $M$ is a positive integer and $\alpha \geq 2$
My attempt:
In my attempt I use the Binomial expansion for $(1+x)^{-M}$ and the result is as follows $$\int_0^{\infty}[Mx-\frac{1}{2}M(M+1)x^2-\frac{1}{6}M(M+1)(M+2)x^3+\cdots]x^{-\frac{2}{\alpha}-1}dx$$ As can be seen that the result will come out to be zero because $\infty^{-j}=0$ and similarly $0^i=0$. Your help in solving this integration will be much appreciated. Thanks in advance.
You correctly pointed out the problem if the binomial expansion is used.
If fact, you need to use hypergeometric functions since $$\int\left(1-\frac{1}{(1+x)^M}\right)x^{-\frac{2}{\alpha}-1}\,dx=\frac{1}{2} \alpha x^{-2/\alpha} \left(\, _2F_1\left(-\frac{2}{\alpha},M;1-\frac{2}{\alpha};-x\right)-1\right)$$ Using a CAS, what was obtained is $$\int_0^\infty\left(1-\frac{1}{(1+x)^M}\right)x^{-\frac{2}{\alpha}-1}\,dx=-\frac{\Gamma \left(-\frac{2}{\alpha}\right) \Gamma \left(M+\frac{2}{\alpha}\right)}{\Gamma (M)}$$