Integral with singularity in $0$ I'm trying to calculate the result of this integral:
$$\int_{0}^{\infty} \mbox{d}x \frac{\sin(\omega x)}{x^{3/2}}$$
where $\omega > 0$, and I know the result is $\sqrt{2 \pi \omega}$, but I have no idea how to come there.
Since there is a singularity in $0$, I can't use Jordan's lemma, that could have been good since the integrand goes to $0$ uniformly for $|z| \to \infty$ in the upper half-plane ($\int_{0}^{\infty} = \frac{1+i}{2} \int_{- \infty}^{\infty}$).
I also tried to trasform the integral in a sum of integrals over every period:
$$\int_{0}^{\infty} \mbox{d}x \frac{\sin(\omega x)}{x^{3/2}} = \sum_{n=0}^{\infty} \int_{2n \pi / \omega}^{2(n+1) \pi / \omega} \mbox{d}x \frac{\sin(\omega x)}{x^{3/2}} =$$
$$[z = \frac{t + 2n \pi}{\omega}, \mbox{d}z = \frac{1}{\omega} \mbox{d}t, \frac{2n \pi}{\omega} \to 0, \frac{2(n+1) \pi}{\omega} \to 2 \pi] $$
$$= \frac{\sqrt{\omega}}{2i} \sum_{n=0}^{\infty} \int_{0}^{2 \pi} \mbox{d}t \frac{e^{it} - e^{-it}}{(t + 2n \pi)^{3/2}} =$$
$$[w = e^{it}, t = -i \ln(w), \mbox{d}t = - \frac{i}{w} \mbox{d}w]$$
$$= - \frac{\sqrt{\omega}}{2} \sum_{n=0}^{\infty} \oint_{|w|=1} \mbox{d}w \frac{1 + 1/w^2}{(2n \pi - i \ln(w))^{3/2}} =$$
[Since the only singularities are in 0 (removable) and 1 (pole), for $n=0$; for $n>0$ all integrals are $=0$]
$$= \sqrt{\frac{\omega}{2}} (1+i) \oint_{|w|=1} \mbox{d}w \frac{w^2 + 1}{w^2 \ln^{3/2} (w)} $$
but there is still a singularity on the path, so I can't calculate the residue and solve the integral.
I suspect that my approach is wrong somewhere...
 A: Let's make a substitution $\omega x = y^2$ where $y=\sqrt{\omega x}$. Then $2 y \mathrm{d} y = \omega \mathrm{d} x$ :
$$
  \int_0^\infty \frac{\sin(\omega x)}{x^{3/2}} \mathrm{d} x = \int_0^\infty \frac{\sin(y^2)}{y^3 \omega^{-3/2}} \, \frac{2 y}{\omega} \mathrm{d} y =2 \sqrt{\omega} \int_0^\infty \frac{\sin(y^2)}{y^2} \, \mathrm{d} y = 2 \sqrt{\omega} \mathcal{I}
$$
It remains to evaluate the integral $\mathcal{I}$, which is now non-singular and does not depend on $\omega$. 
Integration by parts gives:
$$ \begin{eqnarray}
  \mathcal{I} &=& \int_0^\infty \sin(y^2) \mathrm{d} \left( \frac{-1}{y} \right) =
    \left.-\frac{\sin(y^2)}{y} \right|_0^\infty + \int_0^\infty \frac{2 y \cos(y^2)}{y}  \mathrm{d} y \\
   &=& 0 + 2 \int_0^\infty \cos(y^2) \mathrm{d}y = 2 \operatorname{Re}\left( \int_0^\infty \mathrm{e}^{i y^2} \mathrm{d} y\right) = 2 \operatorname{Re}\left( \mathrm{e}^{i \pi/4} \int_0^\infty \mathrm{e}^{-y^2} \mathrm{d} y\right) = \sqrt{\frac{\pi}{2}} 
\end{eqnarray} 
$$
The last equality is obtained by rotating the integration path to ray at $\arg(y) = \frac{\pi}{4}$. Consider $\int_0^R \mathrm{e}^{i y^2} \mathrm{d} y$. Add to it the integration over the counter-clock-wise arc of radius R and the integration along the ray $\arg(z) = \frac{\pi}{4}$. The integration over so obtained closed contour is zero. Integral over arc vanishes as $R \to +\infty$. We have $\int_0^\infty \mathrm{e}^{i (\exp(i \pi/4) x)^2} \mathrm{d} (\exp(i \pi/4) x) = \exp(i \pi/4) \int_0^\infty \exp(-x^2) \mathrm{d} x = \exp(i \pi/4) \frac{\sqrt{\pi}}{2}$.
Combining results:
$$
  \int_0^\infty \frac{\sin(\omega x)}{x^{3/2}} \mathrm{d} x = \sqrt{2 \pi \omega} 
$$
It remains to show that the limit vanishes:
$$
   \left.-\frac{\sin(y^2)}{y} \right|_0^\infty = -\lim_{y \to \infty} \frac{\sin(y^2)}{y} + \lim_{y \to 0} \frac{\sin(y^2)}{y}
$$
The first one does by sandwiching argument $-1 \le \sin(y^2) \le 1$, thus $-\frac{1}{y} \le \frac{\sin(y^2)}{y} \le \frac{1}{y}$, and both lower and upper bound have zero limit.
The latter limit is zero because $ \lim_{y \to 0} \frac{\sin(y^2)}{y} = \lim_{y \to 0} \frac{\sin(y^2)}{y^2} \cdot y = \lim_{y \to 0} \frac{\sin(y^2)}{y^2} \cdot \lim_{y \to 0}  y = 1 \cdot 0$.
