Tricky Indented Contour Integral of $\int_{0}^{\infty} \frac{x^{3/4}}{(x^2+1)^2} \ dx$

Question

I am trying to evaluate the integral $$\int_{0}^{\infty} \frac{x^{3/4}}{(x^2+1)^2} \ dx,$$ by finding a suitable branch of $z^{3/4}$ and integrating the function $z\rightarrow z^{3/4}/(z^2+1)^2$ around the contour below:

$\hskip2in$

I am unsure of how to deal with the branch cut. I know that $w^{1/n}$ is not differentiable for $(-\infty,0]$, but I am unsure what to do, as the contour above includes the negative real axis.

So far, I have show by the ML lemma that for $r<1<R$, $$\int_{\Gamma_1} f(z) \ dz\rightarrow 0 \ \text{as} \ R\rightarrow\infty.$$ Similarly, $$\int_{\Gamma_2} f(z) \ dz\rightarrow 0 \ \text{as} \ r\rightarrow 0.$$ I am having trouble solving the line segments. For $0<\delta<\frac{\pi}{2}$, I have shown that as $\delta\rightarrow 0$, the sum of both line segments equals $0$... but this cannot be correct.

Any advice is greatly appreciated.

Answer 1

Take the branch cut as the positive real axis. Then you have a branch of $z^{3/4}$ given by $re^{it}\mapsto r^{3/4}e^{3it/4}$ for $0\le t<2\pi$. By continuity, on the "lower" horizontal segment of your contour $x^{3/4}$ becomes $x^{3/4}e^{3\pi i/2}=-ix^{3/4}$.

Alternatively avoid the branch cut by making a preliminary substitution $x=y^4$ or $x=e^y$.

Answer 2

$z^{\frac{3}{4}}=\exp(\frac{3}{4}\ln z)$

Below x-axis $\int_\infty ^0\frac{\exp(\frac{3}{4}\ln x+2\pi i)}{1+x^2}\mathrm dx=\int_0^\infty\frac{ix^{\frac{3}{4}}}{1+x^2}\mathrm dx$

So sum of the two path parallel to the x-axis is $(1+i)\int_0^\infty\frac{x^{\frac{3}{4}}}{1+x^2}\mathrm dx$