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Show that for $0<a<1$

$$\int_{0}^{\infty}\frac{x^{a}}{x(x+1)}~\mathrm{d}x=\frac{\pi}{\sin(\pi a)}$$

I want to solve this question by using complex analysis tools but I even don't know how to start. Any help would be great.

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  • $\begingroup$ Shouldn't the right hand side be $\frac{\pi\alpha}{\sin(\pi\alpha)}$? $\endgroup$ – detnvvp Aug 20 '15 at 23:50
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Set $x^{\alpha}=u$, then $du=x^{\alpha-1}\,dx$, and $$\int_0^{\infty}\frac{x^{\alpha}}{x(x+1)}\,dx=\int_0^{\infty}\frac{1}{u^{1/\alpha}+1}\,du.$$ You can then proceed as in this answer.

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    $\begingroup$ Just a remark for the OP. Be careful if $\frac1\alpha$ is not an integer. You can remedy this situation by choosing a branch of $z\mapsto z^{1/\alpha}$, say $z^{1/\alpha}=r^{1/\alpha}\exp(\text{i}\theta/\alpha)$ for $z=r\exp(\text{i}\theta)$ with $\theta\in[0,2\alpha\pi)$ and $r\geq 0$. $\endgroup$ – Batominovski Aug 21 '15 at 0:10
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Here is another straightforward way to proceed that seems easier than the method referenced in another answer herein. So, away we go ...

Let's analyze the contour integral

$$I(a)=\oint_C \frac{z^a}{z(z+1)}\,dx$$

where $C$ is the classical keyhole contour.


NOTE:

The contribution from integrating around the branch point can be shown to go to zero as the radius of that circular part of the contour that "detours around " the branch point tends to zero.


Then, from the residue theorem, we have for $1>a>0$

$$\int_0^{\infty}\frac{x^a}{x(x+1)}\,dx+\int_{\infty}^0\frac{e^{i2\pi a}x^a}{x(x+1)}\,dx=2\pi i \left(\frac{e^{i\pi a}}{-1}\right)$$

and therefore after simplifying $2\pi i \left(\frac{e^{i\pi a}}{e^{i2\pi a}-1}\right)$ we obtain

$$\bbox[5px,border:2px solid #C0A000]{\int_0^{\infty}\frac{x^a}{x(x+1)}\,dx=\frac{\pi}{\sin \pi a}}$$

as expected!!

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I know it's not technically a complex analysis route, but where would we be without the obligatory beta function route?

Rewrite

$$\begin{align}\int_0^{\infty} \frac{x^a}{x (x+1)} &= \int_0^{\infty} \frac{x^{a-1}}{x+1}\\ &= \int_1^{\infty} dx \frac{(x-1)^{a-1}}{x} \\ &= \int_0^1 \frac{dy}{y} \left (\frac1{y}-1 \right )^{a-1} \\ &= \int_0^1 dv \, v^{-a} (1-v)^{-(1-a)}\\ &= \frac{\Gamma(1-a) \Gamma(a)}{\Gamma(1)} \\ &= \frac{\pi}{\sin{\pi a}}\end{align}$$

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  • $\begingroup$ artful! and nice!+1 $\endgroup$ – Math-fun Aug 21 '15 at 19:44

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