Contour integration troubles: Example $\int_0^\infty \frac{x^{1/3}}{1+x^2}$ I know this has been answered on the site (see here: How to integrate $\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$?) but I want to know why what I did is not leading to a sensible result. 
I used the splitting the logarithm trick, i.e. examining 
$$
\int_\Gamma\frac{z^{1/3}\log z}{1+z^2}\mathrm dz=2\pi i\left( \frac{\sqrt{3}}{2}+\frac{i}{2}\right )=\pi(\sqrt{3} i-1)
$$
by the residue theorem and using the sane branch cut on the positive real axis. Now by Jordan's theorem the contribution of the arc should be zero leaving us with 
$$
\int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx+\int_{-R}^0\frac{x^{1/3}\log x}{1+x^2}\mathrm dx\\
=
\int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx+\int_{0}^R\frac{x^{1/3}\log(- x)}{1+x^2}\mathrm dx\\
=2\int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx+\frac{\pi}{2}i\int_0^R\frac{x^{1/3}}{1+x^2}\mathrm dx
$$
which by equating real and imaginary parts should imply my integral is 
$$
2\int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx+\frac{\pi}{2}i\int_0^R\frac{x^{1/3}}{1+x^2}\mathrm dx=\pi(\sqrt{3} i-1)\\
\implies \int_0^\infty\frac{x^{1/3}}{1+x^2}\mathrm dx=2\sqrt{3}
$$
which is false. Additionally, this would imply that 
$$
\int_0^R \frac{x^{1/3}\log x}{1+x^2}\mathrm dx=-\frac{\pi}{2}
$$
which is also false.
Where did I go wrong? My guess is somewhere in the branch cutting, possibly with taking the cube root of 3 in the residue theorem?
 A: To do: Integration along the circle sector $\,\Gamma_R\,$ through the origin 
and limitation by the corner points $\,R>0\,$ and $\,e^{i\pi}R=-R$ .  
The $log$ in your integral is unnecessary and makes the calculation error-prone.
Hint: Wolframalpha tells that $\,\displaystyle\int\limits_0^\infty \frac{x^{1/3}\ln x}{1+x^2}=\frac{\pi^2}{6}\,$ and $\,\displaystyle\int\limits_{-\infty}^0 \frac{x^{1/3}\ln x}{1+x^2}=-\frac{\pi^2}{12}(5-i3\sqrt{3})$ .
With other words: $\,\displaystyle\int\limits_\Gamma \frac{z^{1/3}\ln z}{1+z^2}dz\,$ contents $\,\pi^2\,$, not only $\,\pi\,$. (I haven't calculated this integral but obviously you have a wrong result for this).

Only to see, that it's easier to calculate without any tricks.
We get on the one hand 
$\displaystyle \lim\limits_{R\to\infty} \oint_{\Gamma_R}f dz=i2\pi \cdot\text{res}\left( \frac{z^{1/3}}{1+z^2},z=i \right)= -i\pi e^{i2\pi/3} $ 
and on the other hand 
$\displaystyle \lim\limits_{R\to\infty} \oint_{\Gamma_R}f dz=\int_{\mathbb{R}^+}f dz -\int_{e^{i\pi} \mathbb{R}^+}f dz=\int_{\mathbb{R}^+}f(z) dz -\int_{\mathbb{R}^+}f(e^{i\pi} z) e^{i\pi}  dz$
$\displaystyle =\int_{\mathbb{R}^+}f(z) dz - e^{i2\pi(2/3)} \int_{\mathbb{R}^+}f(z)dz= 
(1- e^{i4\pi/3}) \int_{\mathbb{R}^+}f(z) dz $ .
It follows: 
$\displaystyle \int_{\mathbb{R}^+}f(z) dz =\frac{-i\pi e^{i2\pi/3} }{1-e^{i4\pi/3}}=\frac{\pi}{\sqrt{3}}$
