Integrate using Cauchy Integral Theorem Evaluate the integral
I=$\int_0^\infty \sin(x^3)dx$ 
I already know that the answer is $(1/2)\Gamma(4/3)$. So far I have considered the integral $\int_0^\infty e^{-x^3}dx=\Gamma(4/3)$ which I have already shown to be true, I'm not really sure how to proceed. I am told to consider the integral $\oint_C e^{-z^3}dz$ along the contour C, being simply the contour in the first quadrant. Unfortunately, this is from a math methods for physics course, basically the professor just rambles and writes out sparse solutions to everything on the board (including this one) but I don't really get it since I've never studied complex variables at all before and he never bothered to really go through it (it's not a prerequisite for the course).
 A: Consider the contour integral $\int_{C(R)} e^{-z^3}\, dz$, where $C(R)$ is boundary of a sector of the quarter circle of radius $R$ in the first quadrant (centered at the origin), subtended by the angle $\pi/6$, and the orientation is counterclockwise. Show that the integral of $e^{-z^3}$ along the arc of the sector tends to $0$ as $R\to \infty$. 
Since $e^{-z^3}$ is entire, $\int_{C(R)} e^{-z^3}\, dz = 0$ by Cauchy's theorem. As
$$\int_{C(R)} e^{-z^3}\, dz = \int_0^R e^{-r^3}\, dr + \int_{\text{Arc}} e^{-z^3}\, dz - \int_0^{R} e^{-(re^{i\pi/6})^3}\, e^{i\pi/6}\, dr,$$
taking the limit as $R\to \infty$ yields
$$0 = \int_0^\infty e^{-r^3}\, dx -\int_0^\infty e^{-(re^{i\pi/6})^3}e^{i\pi/6}\, dz$$
or
$$\int_0^\infty e^{-i(r^3-i\pi/6)}\, dr = \int_0^\infty e^{-r^3}\, dr$$
Using Euler's identity $e^{i\theta} = \cos \theta + i\sin \theta$, we can write
$$\int_0^\infty [\cos(r^3 - \pi/6) + i\sin(r^3 - \pi/6)]\, dr = \int_0^\infty e^{-r^3}\, dr.$$
Thus
$$\int_0^\infty \left[\frac{\sqrt{3}}{2}\cos(r^3) + \frac{1}{{2}}\sin(r^3)\right]\, dr + i\int_0^\infty \left[-\frac{1}{2}\cos(r^3) + \frac{\sqrt{3}}{2}\sin(r^3)\right]\, dr=\Gamma\left(\frac{4}{3}\right), $$
using the subtraction formulas for cosine and sine, as well as the evaluation $\int_0^\infty e^{-r^3}\, dr = \Gamma(4/3)$. Taking real and imaginary parts, we get a system of equations
\begin{align}
\frac{\sqrt{3}}{2}A + \frac{1}{2}B &= \Gamma\left(\frac{4}{3}\right)\\
-\frac{1}{2}A + \frac{\sqrt{3}}{2}B &= 0
\end{align}
where $A = \int_0^\infty \cos(r^3)\, dr$ and $B = \int_0^\infty \sin(r^3)\, dr$. The solution is 
$$(A,B) = \left(\frac{\sqrt{3}}{2}\Gamma\left(\frac{4}{3}\right), \frac{1}{2}\Gamma\left(\frac{4}{3}\right)\right).$$ So in particular,
$$B = \int_0^\infty \sin(r^3)\, dr = \frac{1}{2}\Gamma\left(\frac{4}{3}\right).$$
A: If you want to (more or less) avoid complex integration, you can use the properties of the Laplace transform. We have:
$$\mathcal{L}(\sin t) = \frac{1}{1+s^2},\qquad\mathcal{L}^{-1}\left(s^{-2/3}\right)=\frac{1}{t^{1/3}\Gamma(2/3)},$$
hence:
$$ I = \frac{1}{3}\int_{0}^{+\infty}u^{-2/3}\sin u\,du = \frac{1}{3\,\Gamma(2/3)}\int_{0}^{+\infty}\frac{dv}{v^{1/3}(1+v^2)}=\frac{1}{\Gamma(2/3)}\int_{0}^{+\infty}\frac{z\,dz}{1+z^6}$$
from which:
$$ I = \frac{\pi}{3\sqrt{3}\,\Gamma(2/3)}=\frac{\pi\,\Gamma(4/3)}{\sqrt{3}\,\Gamma(2/3)\Gamma(1/3)}=\color{red}{\frac{1}{2}\,\Gamma(4/3)}=\frac{1}{6}\,\Gamma(1/3)$$
as wanted.
