complex analysis - Integrate $int _0^{infty }frac{sqrt[3]{x}}{x^2+1:}:dx$

So I want to compute $ I = \int _0^{\infty }\frac{\sqrt[3]{x}}{x^2+1\:}\:dx$.

First thing I thought of is that integrals from $0$ to $\infty$ usually take a nicer form when we apply the change $x = e^t$, so I did that:
$$
I = \int _{-\infty }^{\infty }\frac{e^{\frac{4}{3}t}}{e^{2t}+1\:}\:dt
$$

Now, I've seen some other integrals with a form like this one. For the other ones, I computed the contour integral of a rectangle of vertex $[-R, R, R+xi, -R+xi]$, choosing $x$ such that the integrand takes opposite values in $[-R, R]$ and $[-R+xi, R+xi]$.

If I could choose such an $x$, then I would have succeeded, since I can prove that the lateral integrals go to $0$ as $R\to \infty$, and thus the desired integrals would be equal to half the value of the residues enclosed by the rectangle.

But that requires solving the following system:
$$
\begin{cases}
e^{\frac{4}{3}t} = -e^{\frac{4}{3}(t+xi)} \\
e^{2t} = e^{2(t+xi)}
\end{cases} \implies
\begin{cases}
\frac{4}{3}xi+i\pi = 2ki\pi \\

2xi = 2ki\pi
\end{cases} \implies
\begin{cases}
x = \frac{3}{4}(2k+1)\pi \\
x = k\pi
\end{cases}
$$
For $k\in \mathbb{Z}$.

But this system does not have a solution!

The first equation forces $x$ to be $q\pi$ for some $q\not\in\mathbb{Z}$, contradicting the second equation.

From here I am stuck. Can I get a hint?

---

EDIT: I followed @DanielFischer suggestion to use a keyhole contour.

Let $C$ be the keyhole contour formed by a big circle $\Gamma_R$ of radii $R$, a small circle $\gamma_\epsilon$ of radii $\epsilon$ and two segments connecting the two circles surrounding the positive axis, separated by a $\delta$ margin.

Then we have thanks to the estimation lemma that:
$$
|\int_{\Gamma_R}|\le {\sup}_{z\in{\Gamma_R}}{\frac{\sqrt[3]{z}}{z^2+1\:}}\cdot long(\Gamma_R)\sim \frac{R^{1/3}}{R^2}\cdot 2\pi R \to 0
$$

On the other hand, when $\delta \to 0$:
$$

\int_R^\epsilon\frac{\sqrt[3]{z}}{z^2+1\:} =
\int_R^\epsilon\frac{e^{\log{z}/3}}{z^2+1\:} =
\int_R^\epsilon\frac{e^{\frac{\log{|z| + i\arg{z}}}{3}}}{z^2+1\:} =
-e^{-2\pi i/3}\int_\epsilon^R\frac{e^{\frac{\log{|z| + i\arg{z}+2\pi i}}{3}}}{z^2+1\:} =
-e^{-2\pi i/3}\int_\epsilon^R
$$

Thus in the limit:
$$
\int_C = \int_\Gamma + \int_\gamma + \int_\epsilon^R + \int_R^\epsilon =

\int_0^\infty -e^{-2\pi i/3}\int_0^\infty =\\
=(1 -e^{-2\pi i/3})\int_0^\infty =
2\pi i (Res(i)+Res(-i))
$$

The residues can be easily calculated as:
$$
Res(i) = \frac{\sqrt[3]{i}}{2i\:}\\
Res(-i) = \frac{\sqrt[3]{-i}}{-2i\:}
$$

Thus $\int_0^\infty = \frac{2\pi i (\frac{\sqrt[3]{i}}{2i\:}+\frac{\sqrt[3]{-i}}{-2i\:})}{(1-e^{-2\pi i/3})} = \frac{π}{2 \sqrt{3}} + \frac{i π}{2}$... which is not real as it should be.

Along the way I've also assumed that $\int_{\gamma_\epsilon}\to 0$, which seems like the case but I cannot prove it.

Therefore, I ask:

1. How do I prove that $\int_{\gamma_\epsilon}\to 0$?


2. Why is my result wrong?

---

Turns out the minus sign on the exponent was wrong:
$$I = \frac{i \left(e^{\frac{i \pi }{3}}-e^{-\frac{2 i \pi }{3}}
\right) \pi }{1-e^{\frac{2
i \pi }{3}}} = -2\pi / \sqrt{3}$$

This result is still wrong according to the almighty Wolfram Alpha, but at least it has a similar form! I'll keep debugging.

Answer

So, after the change of variable $x=e^t$, you have to compute the integral

$$I=\int_{-\infty}^\infty\frac{e^{4/3 t}}{e^{2t}+1}\,dt.$$

Consider the contour
$$\Gamma:=\underbrace{[-R,R]}_{\gamma_1}\cup\underbrace{[R,R+i\pi]}_{\gamma_2}
\cup\underbrace{[R+i\pi,-R+i\pi]}_{\gamma_3}\cup
\underbrace{[-R+i\pi,-R]}_{\gamma_4},$$
and note that $f(z):=\frac{e^{4/3 z}}{e^{2z}+1}$ has only a simple pole at $z=i\pi/2$ inside $\Gamma$.

It is easy to see that the integral of $f$ along $\gamma_2$ and $\gamma_4$ goes to zero, so taking the limit $R\to\infty$ we have

$$I+\lim_{R\to\infty}\int_{\gamma_3}f(z)\,dz=2\pi i\;\text{res}(f,z=i\pi)=-i\pi e^{2i\pi/3}.$$

Now,

$$\int_{\gamma_3}f(z)\,dz=-\int_{-R}^R\frac{e^{4/3(t+i\pi)}}{1+e^{2(t+i\pi)}}\,dt=-e^{4/3i\pi}\int_{\gamma_1}f(z)\,dz$$

so finally

$$I-e^{4/3i\pi}I=-i\pi e^{2i\pi/3}\implies I=\frac12\frac{\pi}{\sin(2\pi/3)}=\frac{\pi}{\sqrt{3}}.$$