Integral Convergence $\sin{x}/x^{3/2}$ Does the following integral converge:
$$\int_0^\infty{\frac{\sin x}{x^{3/2}}}dx$$
I have tried to integrate this by parts and arrived at:
$$-x^{-3/2}\cos x -\int \frac 12{x^{-1/2}}\cos{x} dx $$
Which equals:
$$-x^{-3/2}\cos x -\int \cos{u^2}du$$
I then tried to use the Freshnel Integral for Cos however arrived at an answer that wasn't anywhere close to what Wolfram Alpha provided which was:
$\sqrt{2}\pi$ 
If anyone could help with this problem it would be great.
 A: If $0<x<1$ we have
$$
0<\frac{\sin x}{x^{3/2}}< \frac{1}{x^{1/2}}\quad\text{and}\quad\int_0^1\frac{dx}{x^{1/2}}<\infty\text{ since }1/2<1.
$$
On the other hand
$$
\Bigl|\frac{\sin x}{x^{3/2}}\Bigr|\le\frac{1}{x^{3/2}}\quad\text{and}\quad\int_1^\infty\frac{dx}{x^{3/2}}<\infty\text{ since }3/2>1.
$$
A: The integral converges. First of all
$${\sin(x)\over x^p}\sim {x\over x^p} = {1\over x^{p-1}}$$
as $x\to 0$;
this integrates at zero if and only if $p - 1 < 1$ or if $p  <2$.  Your integrand in fact has an integrable discontinuity there.  Disregard the first two humps so we don't have to do any work there. After that the rest decrease in area, bobbing above and below the $x$-axis.  Now you can see that the  integral converges by the alternating series test.
A: $\frac{\sin x}{x^{3/2}}$ has an integrable singularity in a right neighbourhood of the origin, where $\sin x\sim x$.
On the other hand, Dirichlet's test gives that $\frac{\sin x}{x^{3/2}}$ is improperly Riemann integrable over $[1,+\infty)$, since $\sin x$ has a bounded primitive and $\frac{1}{x^{3/2}}$ decreases to zero on that interval.
Moreover, since $\mathcal{L}(\sin x)=\frac{1}{1+s^2}$ and $\mathcal{L}^{-1}(x^{-3/2})=2\sqrt{\frac{s}{\pi}}$,
$$ \int_{0}^{+\infty}\frac{\sin x}{x^{3/2}}\,dx = \frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{\sqrt{s}}{1+s^2}\,ds=\frac{4}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{t^2}{1+t^4}\,dt=\color{red}{\sqrt{2\pi}}.$$
