How to evaluate the Fourier sine transform of $1/x^3$

Question

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.

Answer 1

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 $F(\omega)$.

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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

$$\sqrt{\frac2\pi}\int_0^\infty \frac{\sin(\omega x)}{x^3}\,dx=-\frac{\sqrt{2\pi}}{4}\omega^2\text{sgn}(\omega)$$

Answer 2

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.

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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 $$ \mathcal{Sine}\{\frac{1}{x^3}\} = \frac{i}{2} \mathcal{F}\{\frac12 \left(\operatorname{pv}\frac{1}{x}\right)''\} = \frac{i}{2} \frac{1}{2} (i\xi)^2 \mathcal{F}\{\operatorname{pv}\frac{1}{x}\} \\ = \frac{i}{2} \frac{1}{2} i^2 \, \xi^2 (-i\pi \operatorname{sign}(\xi)) = -\frac{\pi}{4} \xi^2 \operatorname{sign}(\xi) . $$