How to evaluate the integral $\int \limits_0^\infty \frac{\sin x}{x^{1/3}}\,\mathrm{d}x$? Is there any special convergence factor like $\frac{1}{x}\to\int \limits_0^\infty e^{-xy}\mathrm{d}y$ that can transform the $\frac{1}{x^{1/3}}$? I don't mean $\int \limits_0^\infty e^{-x^{1/3}y}\mathrm{d}y$.
 A: Use the contour $\gamma=[0,R]\cup Re^{i[0,\pi/2]}\cup[iR,0]$ as $R\to\infty$. There are no singularities inside $\gamma$ so
$$
\int_\gamma e^{iz}\,z^{-1/3}\,\mathrm{d}z=0
$$
Furthermore, setting $z=Re^{it}$, we get
$$
\begin{align}
\left|\,\int_{Re^{i[0,\pi/2]}} e^{iz}\,z^{-1/3}\,\mathrm{d}z\,\right|
&\le\int_0^{\pi/2}e^{-R\sin(t)}R^{2/3}\,\mathrm{d}t\\
&\le\int_0^{\pi/2}e^{-2Rt/\pi}R^{2/3}\,\mathrm{d}t\\
&\le R^{2/3}\int_0^\infty e^{-2Rt/\pi}\,\mathrm{d}t\\[6pt]
&=\frac\pi2R^{-1/3}
\end{align}
$$
which tends to $0$ as $R\to\infty$.
Therefore, since the integral along the curve at infinity vanishes, we get
$$\newcommand{\Im}{\operatorname{Im}}
\begin{align}
\int_0^\infty\sin(x)\,x^{-1/3}\,\mathrm{d}x
&=\Im\left(\int_0^\infty e^{ix}\,x^{-1/3}\,\mathrm{d}x\right)\\
&=\Im\left(\int_\gamma e^{iz}\,z^{-1/3}\,\mathrm{d}z+\int_0^{i\infty}e^{ix}\,x^{-1/3}\,\mathrm{d}x\right)\\
&=\Im\left(0+e^{i\pi/3}\int_0^\infty e^{-x}\,x^{-1/3}\,\mathrm{d}x\right)\\[3pt]
&=\sin(\pi/3)\,\Gamma\!\left(\frac23\right)\\[3pt]
&=\frac{\sqrt3}2\,\Gamma\!\left(\frac23\right)
\end{align}
$$
A: In general, for $a$, $s>0$ we have
$$\int_{0}^{\infty}e^{- a x} t^s \frac{ d x}{x}= \frac{\Gamma(s)}{a^s}
$$
This can be extended to $a$,$s \in \mathbb{C}$, $\Re a (,\Re s) >0$ by considering 
$a^s = e^{s \log a}$, with $\log$ the principal branch on the right half plane.  Therefore we have
$$\int_{0}^{\infty} e^{-(t+i)x} x^{\frac{2}{3}}\frac{dx}{x}= \frac{\Gamma(\frac{2}{3})}{\exp(\frac{2}{3}\log(t+i))}$$
As $t\to 0_{+}$, LHS converges to $\int_{0}^{\infty}(\cos x -i \sin x) x^{-\frac{1}{3}} d x$ ( this has to be analyzed with some care), while the RHS clearly converges to $\frac{\Gamma(\frac{2}{3})}{\exp(\frac{2}{3}\cdot i \frac{\pi}{2})}=\Gamma(\frac{2}{3})(\cos \frac{\pi}{3} -i \sin \frac{\pi}{3})$
