Prove $\int_{\mathbb{R^{+}}} \frac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\frac{\pi}{32} (3\pi-4)^2$ How do you arrive at the result $$I=\displaystyle\int_{\mathbb{R^{+}}} \dfrac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\dfrac{\pi}{32} (3\pi-4)^2\ ?$$
Wolfram Alpha agrees numerically.
I tried replacing $\sin p$ by $\dfrac{e^{ip}-e^{-ip}}{2i}$ and similarly for $\cos p$, but in vain.
 A: Here I present another way to solve/generalize the problem. It is much easier. Define
$$ I_q(p)=\frac{1}{2}\int_0^\infty\frac{\sin^3(pt)\cos(qt)}{t^3}dt,p\ge0,q\ge0. $$
Clearly $I_q(0)=I_q'(0)=I_q''(0)=0$ and
\begin{eqnarray}
I_q''(p)&=&\frac{3}{8}\int_0^\infty\frac{(3\sin(3pt)-\sin(pt))\cos(qt)}{t}dt\\
&=&\frac{3}{16}\int_0^\infty\frac{3\sin((3p+q)t)+3\sin((3p-q)t)-\sin((p+q)t)-\sin((p-q)t)}{t}dt\\
&=&\left\{\begin{array}{l}
0, \text{ if }p<\frac{q}{3},\\
\frac{9\pi}{32}, \text{ if }p=\frac q3,\\
\frac{9\pi}{16}, \text{ if }\frac{q}{3}<p<q,\\
\frac{15\pi}{32}, \text{ if }p=q,\\
\frac{3\pi}{8}, \text{ if }p>q.
\end{array}\right.
\end{eqnarray}
Here we used 
$$ \int_0^\infty\frac{\sin(\alpha x)}{x}dx=\text{sgn}(\alpha)\frac{\pi}{2}. $$
Thus
$$ I_q'(p)=\left\{\begin{array}{l}
0, \text{ if }p\le\frac{q}{3},\\
\frac{9\pi}{16}(p-\frac{q}{3}), \text{ if }\frac{q}{3}<p\le q,\\
\frac{3p\pi}{8}, \text{ if }p>q,
\end{array}\right. $$
and hence
$$ I_q(p)=\left\{\begin{array}{l}
0, \text{ if }p\le\frac{q}{3},\\
\frac{\pi}{32}(3p-q)^2, \text{ if }\frac{q}{3}<p\le q,\\
\frac{(3p^2-q^2)\pi}{16}, \text{ if }p>q.
\end{array}\right. $$
For $p=\pi,q=4$, since $\frac{q}{3}<p<q$, we have
$$ I_4(\pi)=\frac{\pi}{32}(3\pi-4)^2.$$
A: The following is an approach using contour integration.
$ $
$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{3}(\pi x^{2}) \cos(4x^{2})}{x^{5}} \ dx &= \frac{1}{2}\int_0^{\infty} \frac{\sin^3(\pi t)\cos(4t)}{t^3} \ dt \\ &= \frac{1}{4} \int_{-\infty}^{\infty} \frac{\sin^{3} (\pi t) \cos(4t)}{t^{3}} \ dt  \\ &= \frac{1}{4} \text{Re} \int_{-\infty}^{\infty} \frac{\sin^{3} (\pi t) e^{4it}}{t^{3}} \ dt \\ &= \frac{1}{16} \text{Re} \int_{-\infty}^{\infty} \frac{[3 \sin (\pi t) - \sin(3\pi t)]e^{4it}}{t^{3}} \ dt\\ &= \frac{1}{32}  \text{Re} \frac{1}{i} \int_{-\infty}^{\infty} \frac{(3e^{\pi i t} - 3e^{ -\pi i t} - e^{3 \pi i t} +  e^{-3 \pi i t})e^{4it}}{t^{3}} \ dt \\ &= \frac{1}{32} \text{Re} \frac{1}{i} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{3e^{(4+\pi) i t} - 3 e^{(4- \pi) i t} - e^{(4+3 \pi) i t} +  e^{(4-3 \pi) i t}}{(t-i \epsilon)^{3}} \ dt \\ &= \frac{1}{32} \text{Re} \frac{1}{i} \lim_{\epsilon \to 0^{+}} 2 \pi i \  \text{Res} \left[ \frac{3e^{(4+\pi)iz} - 3 e^{(4- \pi)iz} - e^{(4+ 3 \pi)iz}}{(z- i \epsilon)^{3}},i \epsilon \right] \\ &= \frac{\pi }{16} \text{Re} \lim_{\epsilon \to 0^{+}} \frac{1}{2!} \lim_{z \to i \epsilon} \frac{d^{2}}{dz^{2}} \left( 3e^{(4+ \pi)iz} - 3 e^{(4-\pi)iz} - e^{(4+ 3\pi)iz}\right) \\ &= \frac{\pi}{32} \left(-3( 4+ \pi)^{2} + 3 (4- \pi)^{2} +(4 + 3 \pi)^{2}\right) \\ &=  \frac{\pi}{32} \left(3 \pi - 4 \right)^{2} \end{align}$$
EDIT:
Notice that on line 6, $4 - 3 \pi <0$. 
So by Jordan's lemma, $\displaystyle  \int_{-\infty}^{\infty}\frac{e^{(4- 3\pi)it}}{(z-i \epsilon)^{3}} dt$ would need to be broken off and evaluated separately by closing the contour with the lower half of $|z|=R$ as opposed to the upper half of $|z|=R$. 
But since the pole is in the upper half-plane, $\displaystyle \int_{-\infty}^{\infty} \frac{e^{(4- 3\pi)it}}{(t-i \epsilon)^{3}} \ dt $ evaluates to $0$.
