calculus - Evaluating a parametric integral

I need some help to evaluate the following integral.

$$\int_{0}^{\infty}{\mathrm{d}x \over x^{\alpha}\left(x + 1\right)}$$ where $\alpha \in \left(0,1\right)$

I've tried many ways (the best one seems to be developing by Taylor series) but actually I have no solution.

Some ideas?
Thank you.

Answer

I think we will need some complex analysis here.

Take a branch of $1/z^a$ defined in $\mathbb{C}\setminus [0,+\infty)$ and consider the integral

$$\int_{\gamma}\frac{dz}{z^a(z+1)}$$
where $\gamma$ is a close path composed by an arc of a inner circle of radius $01$and two parallel segments over and under the segment$[r,R]$. Then by the residue theorem$$\int_{\gamma}\frac{dz}{z^a(z+1)}=2\pi i\mbox{Res}\left(\frac{1}{z^a(z+1) },-1\right)=2\pi i e^{-i\pi a}.$$Now we take the limit as$R\to+\infty$and$r\to 0^+$. It is easy to see that the integrals along the arcs of the circles goes to$0$. Hence$$\int_0^{+\infty}\frac{dx}{x^a(x+1)}-\int_0^{+\infty}\frac{dx}{x^ae^{2\pi ia}(x+1)}=2\pi i e^{-i\pi a}$$which implies that$$\int_0^{+\infty}\frac{dx}{x^a(x+1)}=\frac{2\pi i e^{-i\pi a}}{1-e^{-2i\pi a}}=\frac{\pi}{\sin (\pi a)}.$$