Indented Path Integration The goal is to show that 
$$\int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx = \frac{\pi^2}{6}$$
and that
$$\int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx = \frac{\pi}{\sqrt{3}}.$$
So, we start with the function
$$f(z) = \frac{z^{1/3}\log(z)}{z^2 + 1}.$$
Let $c_r$ be the upper semicircle with radius r such that $r< 1$ and let $c_R$ be the upper semicircle with radius R such that $R>1$. Let $L_1$ be the interval $(r,R)$ on the reals and let $L_2$ be the interval $(-R,r)$ on the reals. 
Then
$$\int_{c_r} f + \int_{c_R} f + \int_{L_1} f + \int_{L_2} f$$
taken counterclockwise is equal to
$$2\pi i\text{ res}(f,i) = -\frac{\pi^2  e^{\frac{\pi i}{6}}}{2}$$
The integrals along $c_r$ and $c_R$ should go to $0$ as $r$ goes to $0$ and $R$ goes to infinity. 
So:
$$\int_{L_1} + \int_{L_2}= \int_0^{\infty}\frac{x^{1/3}\log(x)}{x^2 + 1}dx - \int_{0}^{\infty} \frac{x^{1/3}e^{i\pi/3}(ln(x) + i\pi) }{x^2 + 1}dx$$
where x is the polar radius. 
Let $A =  \int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx$
and let $B = \int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx $.
Then we have $A*(1-e^{\frac{i\pi}{3}}) - \pi i e^{\frac{i\pi}{3}}B =  $-$\frac{\pi^2  e^{\frac{\pi i}{6}}}{2}$.
This doesn't mesh with the given values for $A$ and $B$, so I'm wondering where I went wrong. Thanks in advance. 

Indented contours pick up half a residue. So the integral along $c_r$ is not zero.
