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 $0
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
$$\int_0^{+\infty}\frac{dx}{x^a(x+1)}=\frac{2\pi i e^{-i\pi a}}{1-e^{-2i\pi a}}=\frac{\pi}{\sin (\pi a)}.$$
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