How to evaluate the Fourier sine transform of $1/x^3$ With the help of Maple, I have get the Fourier sine transform of $1/x^3,$ which is defined as $\sqrt{\frac{2}{\pi}}\int_0^{+\infty}\frac{\sin(x\omega)}{x^3}d x.$
And  the output Maple given is
with(inttrans):
fouriersin(1/x^3,x,omega);
-sqrt(2)*sqrt(Pi)*omega^2/4                

But I do not know how Maple calculated this. So I try to evaluate it by hand: By integration by parts,
\begin{align*}
&\sqrt{\frac{2}{\pi}}\int_0^{+\infty}\frac{\sin(x\omega)}{x^3}d x\\
=&\sqrt{\frac{2}{\pi}}\left(\left(\sin(x\omega)\frac{x^{-2}}{-2}-\omega\cos(x\omega)\frac{x^{-1}}{-2(-1)}\right)\bigg|_{x=0}^{x=+\infty}+\int_0^{+\infty}(-1)\omega^2\frac{\sin(x\omega)}{2x}d x\right).
\end{align*}
But it is clear that the function $\sin(x\omega)\frac{x^{-2}}{-2}-\omega\cos(x\omega)\frac{x^{-1}}{-2(-1)}$ is not convergent as $x\to 0$ from right. We known that, by Dirichlet integral, noting $x>0,$
\begin{gather*}
\int_0^{+\infty}\frac{\sin(x\omega)}{x}d x=\frac{\pi}{2}.
\end{gather*}
Thus, if we can show that
\begin{gather*}\tag{$\star$}
\lim_{x\to 0^+}\left(\sin(x\omega)\frac{x^{-2}}{-2}-\omega\cos(x\omega)\frac{x^{-1}}{-2(-1)}\right)=0,
\end{gather*}
then we arrive at $$\sqrt{\frac{2}{\pi}}\int_0^{+\infty}\frac{\sin(x\omega)}{x^3}d x=-\frac{\sqrt{2\pi}\omega^2}{4}.$$
But $(\star)$ is not true. So I think that we can not use ordinary method of integration by parts to this integral. Now what should I do? Or can someone give me some hints or references.
PS: This problem comes out of the 5th exercise on page 323 of Weinberger's book A First Course in Partial Differential Equations with Complex Variables and Transform Methods, 1995.
 A: Rewriting the sine transform as a Fourier transform, $\frac{1}{x^3}$ as $-\frac12(\frac1x)''$ and interpreting $\frac1x$ as the principial value distribution, we get
$$
\int_0^\infty \frac{1}{x^3} \sin \omega x \, dx 
= \frac12 \left( \int_{0}^\infty \frac{1}{x^3} \sin \omega x \, dx 
+ \int_{-\infty}^{0} \frac{1}{x^3} \sin \omega x \, dx \right) 
= \frac12 \int_{-\infty}^\infty \frac{1}{x^3} \sin \omega x \, dx \\
= - \frac12 \operatorname{Im} \int_{-\infty}^\infty \frac{1}{x^3} e^{-i\omega x} \, dx 
= - \frac14 \operatorname{Im} \int_{-\infty}^{\infty} \left(\frac{1}{x}\right)'' e^{-i\omega x} \, dx
= \frac14 \omega^2 \operatorname{Im} \int_{-\infty}^{\infty} \frac{1}{x} e^{-i\omega x} \, dx \\
= \frac14 \omega^2 \operatorname{Im} \left(-i\pi \operatorname{sign}(\omega) \right)
= - \frac{\pi}{4} \omega^2 \operatorname{sign}(\omega)
.
$$
Thus,
$$
\sqrt{\frac{2}{\pi}} \int_0^\infty \frac{1}{x^3} \sin \omega x \, dx
= \sqrt{\frac{2}{\pi}} \left( - \frac{\pi}{4} \omega^2 \operatorname{sign}(\omega) \right)
= - \frac{\sqrt{2\pi}}{4} \omega^2 \operatorname{sign}(\omega)
,
$$
which is the same as Mark Viola got.

Sine transform in terms of Fourier transform
Functions
Let the Fourier transform on $L^1(\mathbb{R})$ be defined by
$$
\mathcal{F}\{f\} := \int_{-\infty}^{\infty} f(x) \, e^{-i\xi x} \, dx
$$
and the sine transform on $L^1(0, \infty)$ be defined by
$$
\mathcal{Sine}\{f\} := \int_0^\infty f(x) \, \sin\xi x \, dx.
$$
Then,
$$\mathcal{Sine}\{f\} = \frac{i}{2} \mathcal{F}\{\bar{f}\},$$
where the oddization $\bar{f}$ of $f$ is defined by
$$
\bar{f}(x) = \begin{cases}
f(x) & (x>0) \\
-f(-x) & (x<0)
\end{cases}
$$
Distributions
Let $u \in \mathcal{S}'(0, \infty)$ have an odd extension $\bar{u} \in \mathcal{S}'(\mathbb{R}).$ Then we define the sine transform of $u$ by
$$
\mathcal{Sine}\{u\} := \frac{i}{2} \mathcal{F}\{\bar{u}\}
$$
The case in the question
With $u = \frac{1}{x^3}$ and $\bar{u} = \frac12 \left(\operatorname{pv}\frac{1}{x}\right)''$ we get
$$
\mathcal{Sine}\{\frac{1}{x^3}\}
= \frac{i}{2} \mathcal{F}\{\frac12 \left(\operatorname{pv}\frac{1}{x}\right)''\}
= \frac{i}{2} \frac{1}{2} (i\xi)^2 \mathcal{F}\{\operatorname{pv}\frac{1}{x}\} \\
= \frac{i}{2} \frac{1}{2} i^2 \, \xi^2 (-i\pi \operatorname{sign}(\xi)) 
= -\frac{\pi}{4} \xi^2 \operatorname{sign}(\xi)
.
$$
A: Let $F(\omega)$ be given by
$$F(\omega)=\sqrt{\frac2\pi}\int_0^\infty \frac{\sin(\omega x)}{x^3}\,dx$$
Clearly, this integral diverges for $\omega \ne0$ due to the sharp singularity at $x=0$.  However, we can give a distributional interpretation to $F(\omega)$.


Denote by $F_\varepsilon(\omega)$ the integral
$$ F_\varepsilon(\omega)=\text{Re}\left(\sqrt{\frac2\pi}\int_0^\infty \frac{\sin(\omega x)}{(x+i\varepsilon)^3}\,dx\right)\tag1$$
Then, application of integration by parts twice to the integral on the right-hand side of $(1)$ reveals
$$\begin{align}
F_\varepsilon(\omega)&=-\frac12\sqrt{\frac2\pi}\text{Re}\left(\int_0^\infty \frac{\omega^2\sin(\omega x)}{x+i\varepsilon}\,dx\right)\tag2
\end{align}$$
Letting $\varepsilon\to 0$ in $(2)$, we find that
$$\lim_{\varepsilon\to0}F_\varepsilon(\omega)=-\frac{\sqrt{2\pi}}{4}\text{sgn}(\omega)\omega^2$$
We interpret $F(\omega)$ as the distributional limit of $(2)$.  That is, for a suitable test function $\phi(\omega)$ we have
$$\begin{align}
\langle \phi,F\rangle &=\lim_{\varepsilon \to 0}\int_{-\infty}^\infty \phi(\omega)F\varepsilon(\omega)\,d\omega\\\\
&=\int_{-\infty}^\infty \phi(\omega)\left(-\frac{\sqrt{2\pi}}{4}\omega^2\text{sgn}(\omega)\right)\,d\omega\\\\\
\end{align}$$
Therefore, in distribution
$$\sqrt{\frac2\pi}\int_0^\infty \frac{\sin(\omega x)}{x^3}\,dx=-\frac{\sqrt{2\pi}}{4}\omega^2\text{sgn}(\omega)$$
