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 $[-R,-r]$? 
 A: To first answer the question in the OP regarding the applicability of a branch cut along the negative imaginary axis, we note that we can certainly evaluate the integral of $\frac{z^a \log(z)}{z^2+1}$ over the proposed contour as
$$\begin{align}\oint_C \frac{z^a \log(z)}{z^2+1} &=\int_r^R \frac{x^a\log(x)}{x^2+1}\,dx-\int_{r}^{R}\frac{e^{i\pi a}x^a(\log(x)+i\pi )}{x^2+1}\,dx\\\\
&+\int_{0}^\pi \frac{R^ae^{ia\phi}(\log(R)+i\phi)}{R^2e^{i2\phi}+1}\,iRe^{i\phi}\,d\phi\\\\&-\int_{0}^\pi \frac{r^ae^{ia\phi}(\log(r)+i\phi)}{r^2e^{i2\phi}+1}\,ire^{i\phi}\,d\phi\end{align}$$
Proceeding, for $0<a<1$, the third and fourth integrals vanish as $r\to 0$ and $R\to \infty$ and the first and second become in the limit 
$$(1-e^{i\pi a})\int_0^\infty \frac{x^a\log(x)}{x^2+1}\,dx-i\pi e^{i\pi a}\int_0^\infty \frac{x^a}{x^2+1}\,dx$$
which requires evaluation of two integrals.   In that which follows, we present a different way forward for which we evaluate a single integral only.  To that end, we proceed.

Note that the integral of interest can be expressed as
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^{1/3}\log(x)}{x^2+1}\,dx=\left. \frac{d}{da}\left(\int_0^\infty \frac{x^a}{x^2+1}\,dx\right)\right|_{a=1/3} }\tag 1$$ 

Therefore, we proceed by evaluating the integral $I(a)$ given by
$$I(a)=\int_0^\infty \frac{x^a}{x^2+1}\,dx \tag 2$$
for $0<a<1$.
We move to the complex plane.  Let $C$ be the classical key-hole contour with the keyhole along the positive real axis.  Then, we have 
$$\begin{align}
\oint_C \frac{z^a}{z^2+1}\,dz&=\int_\epsilon^R \frac{x^a}{x^2+1}\,dx+\int_R^\epsilon \frac{e^{i2\pi a}x^a}{x^2+1}\,dx\\\\
&+\int_0^{2\pi}\frac{R^ae^{ia\phi}}{R^2e^{i2\phi}}\,iRe^{i\phi}\,d\phi+\int_{2\pi}^0 \frac{\epsilon e^{ia\phi}}{\epsilon^2 e^{i2\phi}+1}\,i\epsilon^{i\phi}\,d\phi\\\\
&=(1-e^{i2\pi a})\int_\epsilon^R \frac{x^a}{x^2+1}\,dx\\\\
&+\int_0^{2\pi}\frac{R^ae^{ia\phi}}{R^2e^{i2\phi}}\,iRe^{i\phi}\,d\phi-\int_0^{2\pi} \frac{\epsilon e^{ia\phi}}{\epsilon^2 e^{i2\phi}+1}\,i\epsilon^{i\phi}\,d\phi \tag 3
\end{align}$$
As $R\to \infty$ and $\epsilon \to 0$ the second and third integrals on the right-hand side of $(3)$ vanish while the first integral becomes $(1-e^{i2\pi a})I(a)$ as given in $(2)$.  
Next, applying the residue theorem we find that 
$$J(a)=2\pi i \left(\frac{e^{i\pi a/2}}{2i}+\frac{e^{i3\pi a/2}}{-2i}\right)=-2\pi i e^{i\pi a}\sin(\pi a/2) \tag 4$$
Putting together the limit of $(3)$ and the result from $(4)$ reveals
$$I(a)=\frac{\pi}{2\cos(\pi a/2)} $$
from which we find that 
$$I'(1/3)=\frac{\pi^2}{6}$$
Finally, from $(1)$ find that 
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^{1/3}\log(x)}{x^2+1}\,dx=\frac{\pi^2}{6}}$$
A: 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  $R\rightarrow\infty.$
