Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $ The following is an exercisein complex analysis:

Use contour integrals with $-\pi/2<\operatorname{arg} z<3\pi/2$ to compute
  $$
I:=\int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx.
$$

I don't see why the branch $-\pi/2<\operatorname{arg} z<3\pi/2$ would work. Let
$$
f(z)=\frac{z^{1/3}\log z}{z^2+1}.
$$
Denote $\Gamma_r={re^{it}:0\leq t\leq \pi}$. One can then consider a contour consisting of $\Gamma_R$, $\Gamma_r$, and $[-R,-r]\cup[r,R]$. The integral along $[r,R]$ will give $I$. But without symmetries, how could one deal with the integral along $[-R,-r]$? 

Sketch of a  possible argument. Use the branch  with the argument from
zero to $2\pi$ of the logarithm, the function
$$f(z) = \frac{\exp(1/3\log z)\log z}{z^2+1}$$
and a  keyhole contour  with the slot  aligned with the  positive real
axis.
The sum of the residues at $z=\pm i$ is (take care to use the chosen branch of the logarithm in the logarithm as well as in the power/root)
$$\rho_1+\rho_2 = \frac{1}{8}\,\pi \,\sqrt {3}-\frac{5}{8}\,i\pi.$$
Now below the keyhole we get
$$\exp(1/3\log z)\log z
= \exp(2\pi i/3) \exp(1/3\log x)
(\log x + 2\pi i).$$
Introduce the labels
$$J = \int_0^\infty \frac{x^{1/3}\log x}{x^2+1} dx$$
and
$$K = \int_0^\infty \frac{x^{1/3}}{x^2+1} dx$$
The contribution from the line below the slot is
$$-\exp(2\pi i/3) J - 2\pi i\exp(2\pi i/3) K.$$
We evaluate $K$ and obtain
$$K = \pi\frac{\sqrt{3}}{3}.$$
Solving $$2\pi i (\rho_1+\rho_2) =
(1-\exp(2\pi i/3)) J 
- 2\pi i\exp(2\pi i/3) \pi\frac{\sqrt{3}}{3}$$
we obtain
$$J =\frac{\pi^2}{6}.$$
Remark.
The evaluation of $K$ is recursive and uses the same contour.
We get for the two residues
$$\rho_1+\rho_2 = -\frac{1}{4} i\sqrt {3}-\frac{1}{4}$$
and hence we have 
$$2\pi i (\rho_1+\rho_2) =
(1-\exp(2\pi i/3)) K$$
which we may solve for $K$ to get the value cited above.

Remark II.  The fact that the circular  component vanishes follows
by the ML inequality which gives in the case of evaluating $J$
$$2\pi R \times \frac{R^{1/3}\log R}{R^2-1}
\rightarrow\ 0$$
as  $R\rightarrow\infty.$
