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:


*

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

*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.
 A: Let $\sqrt[3]{x}=\sqrt{u}$, then $x^2=u^{3} \implies 2x\, dx=3u^2\, du \implies dx=\dfrac{3}{2} \sqrt{u}\, du$
\begin{align*}
  \int_{0}^{\infty} \frac{\sqrt[3]{x} \, dx}{x^2+1}
  &= \int_{0}^{\infty} \frac{3u\, du}{2(u^3+1)} \\
  &= \int_{0}^{\infty}
     \left[
       \frac{u+1}{2(u^2-u+1)}-\frac{1}{2(u+1)}
     \right] \, du \\
  &= \int_{0}^{\infty}
     \left[
       \frac{(2u-1)+3}{4(u^2-u+1)}-\frac{1}{2(u+1)}
     \right] \, du \\
  &= \int_{0}^{\infty}
     \left[
       \frac{2u-1}{4(u^2-u+1)}+
       \frac{3}{(2u-1)^2+3}-\frac{1}{2(u+1)}
     \right] \, du \\
  &= \left[
       \frac{1}{4} \ln (u^2-u+1)+
       \frac{\sqrt{3}}{2} \tan^{-1} \frac{2u-1}{\sqrt{3}}-
       \frac{1}{2} \ln (u+1)
     \right]_{0}^{\infty} \\
  &= \left[
       \frac{1}{4} \ln \frac{u^2-u+1}{(u+1)^2}+
       \frac{\sqrt{3}}{2} \tan^{-1} \frac{2u-1}{\sqrt{3}}
     \right]_{0}^{\infty} \\
  &= \frac{\pi}{\sqrt{3}}
\end{align*}

By contour integral
  \begin{align*}
  \oint_{C} \frac{\sqrt[3]{x} \, dx}{x^2+1}
  &= \int_{\epsilon}^{R} \frac{\sqrt[3]{x} \, dx}{x^2+1}+
     \int_{0}^{2\pi}
     \frac{\sqrt[3]{R}e^{i\theta/3} Rie^{\theta}\, d\theta}
          {R^2e^{2i\theta}+1} \\
  &\quad
  -\int_{R}^{\epsilon} \frac{\sqrt[3]{x}e^{2i\pi/3} \, dx}{x^2+1}-
  \int_{0}^{2\pi}
  \frac{\sqrt[3]{\epsilon}e^{i\theta/3} \epsilon ie^{\theta}\, d\theta}
       {\epsilon^2e^{2i\theta}+1} \\
  &= \left( 1-e^{2i\pi/3} \right)
     \int_{\epsilon}^{R} \frac{\sqrt[3]{x} \, dx}{x^2+1}+
     \frac{i}{\sqrt[3]{R^2}}
     \int_{0}^{2\pi}
     \frac{e^{-2i\theta/3} \, d\theta}
          {1+\dfrac{e^{-2i\theta}}{R^2}}-
    i\sqrt[3]{\epsilon^{4}} \int_{0}^{2\pi}
    \frac{e^{4i\theta/3}\, d\theta}
         {1+\epsilon^2e^{2i\theta}} \\
  &= \left( 1-e^{2i\pi/3} \right)
     \int_{0}^{\infty} \frac{\sqrt[3]{x} \, dx}{x^2+1} \\
  \int_{0}^{\infty} \frac{\sqrt[3]{x} \, dx}{x^2+1} &=
  \frac{2\pi i}{1-e^{2i\pi/3}}
  \left[
    \operatorname{Res}
    \left( \frac{\sqrt[3]{x}}{x^2+1},  i \right)+
    \operatorname{Res}
    \left( \frac{\sqrt[3]{x}}{x^2+1}, -i  \right)
  \right] \\
  &= \frac{2\pi i}{1-e^{2i\pi/3}}
  \left(
    \frac{\sqrt[3]{i}}{i+i}+\frac{\sqrt[3]{-i}}{-i-i}
  \right) \\
  &= \frac{\pi(e^{i\pi/6}-e^{i\pi/2})}{1-e^{2i\pi/3}} \\
  &= \frac{\pi(e^{-i\pi/6}-e^{i\pi/6})}{e^{-i\pi/3}-e^{i\pi/3}} \\
  &= \frac{\pi \sin \dfrac{\pi}{6}}{\sin \dfrac{\pi}{3}} \\
  &= \frac{\pi}{\sqrt{3}}
\end{align*}

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