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.

  • $\begingroup$ Shouldn't the right hand side be $\frac{\pi\alpha}{\sin(\pi\alpha)}$? $\endgroup$ – detnvvp Aug 20 '15 at 23:50

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.

  • 1
    $\begingroup$ 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

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.


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!!


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


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

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

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