What is the exact value of $\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$ 
I would like  to get the exact value of the following integral.
  $$\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$$

I was able to prove the convergence as well. But I don't how to compute its exact value. I tried with the Residue Theorem of the complex function 
$$z\mapsto \frac{\sin^2 z}{z^{5/2}}$$
But I could not move further. 
 A: With a step of integration by parts the problem boils down to computing 
$$ \frac{2}{3}\int_{0}^{+\infty}\frac{\sin(2x)}{x^{3/2}}\,dx =\frac{2\sqrt{2}}{3}\int_{0}^{+\infty}\frac{\sin x}{x^{3/2}}\,dx\stackrel{\mathcal{L},\mathcal{L}^{-1}}{=}\frac{4\sqrt{2}}{3\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}}{s^2+1}\,ds$$
and the last integral is elementary (just enforce the substitution $s\mapsto u^2$ and perform a partial fraction decomposition). By the properties of the Laplace transform, the final outcome is  $\frac{4}{3}\sqrt{\pi}$.
A: Performing the change of variables $2u = x^2$ together with two integration by parts, we get, $$ \int_0^\infty \cos(x^2)dx =  \frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx\\=\frac{1}{2\sqrt{2}}\underbrace{\left[\frac{\sin 2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{4\sqrt{2}} \int^\infty_0\frac{\sin 2 x}{x^{3/2}}\,dx\\=
 \frac{1}{4\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{3}{8\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$$
Hence $$\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx = \frac{8\sqrt2}{3}\int_0^\infty \cos(x^2)dx = \frac{4\sqrt \pi}{3}$$
Since See Here,  $$\int_0^\infty \cos(x^2)dx= \sqrt\frac\pi8$$
or
How to prove only by Transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $
A: This integral $I$ can be interpreted as the Mellin transform of $\sin(x)^2$ evaluated at $s=-3/2$.
$$
\mathcal{M}[\sin(x)^2] = -2^{-1-s}\cos\left(\frac{\pi s}{2}\right)\Gamma(s)
$$
so
$$
I = \Gamma\left(-\frac{3}{2}\right) = \frac{4\sqrt{\pi}}{3}
$$
