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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.

azc
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  • The integral diverges due to the sharp singularity at $x=0$. – Mark Viola Jun 21 '20 at 03:52
  • @MarkViola: Yes, the integral is divergent according to the classical definition of improper integral. But why Maple gives the result. Maybe it uses a different definition, for instance, by distribution (generalized function). – azc Jun 21 '20 at 04:05
  • https://www.wolframalpha.com/input/?i=fourier+sin+transform+of+%281%2Fx%5E3%29 gives a result as well. – Kraigolas Jun 21 '20 at 04:09
  • @VividKraig: Because MMA and Maple use the same definition of Fourier sine transform. – azc Jun 21 '20 at 04:12
  • @MarkViola: If put this as a distribution, which space of test function should I take? If I take $C^{\infty}_c((0,+\infty)),$ then the desired result follows directly, for the boundary integral vanished. Can you give some references on distribution? – azc Jun 22 '20 at 01:13

2 Answers2

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

md2perpe
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  • I hope you're staying safe and healthy my friend. This development seems like formal manipulation in which the object $\int_0^\infty \frac{\sin(\omega x)}{x^3},dx$ is treated as an integral, although the integral diverges. As you know, if the object is a distribution, it needs to act on a suitable test function. How does that work here? I tried to regularize the distribution, but this led to the issues that we discussed yesterday. And another user has asserted that $\lim_{\varepsilon\to 0}\frac1{x^3}1_{|x|\ge \varepsilon}$ fails to converge in distribution. How can this be made rigorous? – Mark Viola Jun 23 '20 at 16:10
  • Yes, it's first a formal manipulation to reach something that can be treated as a distributional Fourier transform. But the Fourier transform itself, $\mathcal{F}{\frac 1 x}=-i\pi\operatorname{sign}(\omega),$ is rigorous, although I still write it as an integral. – md2perpe Jun 23 '20 at 16:21
  • Calculation of Fourier transform of $\operatorname{pv}\frac1x$ – md2perpe Jun 23 '20 at 16:28
  • Earlier calculation of transform of $\operatorname{pv}\frac{1}{x^3}$ – md2perpe Jun 23 '20 at 16:31
  • Yes, of course I understand that the Fourier Transform of $1/x$ is well defined as a Cauchy principal value. – Mark Viola Jun 23 '20 at 16:52
  • The formal calculations can probably be made rigorous by defining the sine transform of a distribution on $(0,\infty)$ in terms of the Fourier transform through extending the distribution to $(-\infty,\infty).$ The latter requires that the distribution is well-behaved at $0$. – md2perpe Jun 23 '20 at 17:10
  • @MarkViola. I added some more rigorous treatment defining sine transform in terms of Fourier transform and odd extension of the distribution (when such exists). – md2perpe Jun 23 '20 at 18:23
  • I've edited my answer considerably. In particular, I am using a different regularization for $\int_0^\infty \frac{\sin(\omega x)}{x^3},dx$. Please let me know your thoughts. – Mark Viola Jun 23 '20 at 19:27
  • @MarkViola. Sure. Done. – md2perpe Jun 23 '20 at 19:55
  • And (+1) for the solid answer! Well done. – Mark Viola Jun 23 '20 at 20:07
  • @MarkViola. We complement each other. – md2perpe Jun 23 '20 at 20:14
  • Indeed. The two answers arrive at the same conclusion using distinct approaches. I'd be curious to see others. – Mark Viola Jun 23 '20 at 20:15
2

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

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

Mark Viola
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  • You have only shown the limit for $\phi'' \in S,$ not for general $\phi \in S$. You can therefore only conclude equality modulo ${ ax+b \mid a,b \in \mathbb{C} }.$ – md2perpe Jun 22 '20 at 21:30
  • $g(x) = e^{-x^2} \in S$ but no primitive function of $g$ is in $S$ since the primitive cannot tend to $0$ for both $x \to \infty$ and $x \to -\infty.$ – md2perpe Jun 22 '20 at 22:11
  • That's correct, $ax+b$ is NOT a Schwartz space function. But it is a tempered distribution. That's what I meant although it wasn't clear. From your calculations you cannot conclude that $F(\omega) = - \frac{\sqrt{2\pi}}{4} \omega^2 \operatorname{sgn}(\omega)$ but only that $$F(\omega) = - \frac{\sqrt{2\pi}}{4} \omega^2 \operatorname{sgn}(\omega) + ax +b$$ for some $a,b \in \mathbb{C}.$ – md2perpe Jun 22 '20 at 22:15
  • I made an error, it should of course be $ \cdots + a\omega + b.$ – md2perpe Jun 22 '20 at 22:23
  • Take $\phi'' \in S.$ Then $$ \langle x, \phi'' \rangle = - \langle 1, \phi' \rangle = \langle 0, \phi \rangle = 0 = \langle 0, \phi'' \rangle . $$ Can we conclude that $x = 0$ in $S'$? – md2perpe Jun 22 '20 at 22:26
  • $\lim_{\epsilon \to 0} x^{-3}1_{|x|>\epsilon}$ doesn't converge in the sense of distributions. You need first to explain to the OP that we need a distribution agreeing with $x^{-3}$ on $\Bbb{R}^*$, for example $$<T,\phi>=\int_R \frac{\phi(x)-\phi(0)-\phi'(0)x-1_{|x|<1}\phi''(0) x^2 /2}{x^3}dx$$ Then $x^3T=1$, $\hat{T}''' = c\delta$ so $\hat{T}=c \xi^2 1_{\xi > 0}/2+P$ with $P$ a degree 2 polynomial that we deduce from the real oddness of $T$ @md2perpe – reuns Jun 23 '20 at 00:48
  • My comment is clear, we need a distribution, you don't have any distribution. – reuns Jun 23 '20 at 03:57
  • @reuns I've edited considerably and have changed the interpretation of the object $\int_0^\infty \frac{\sin(\omega x)}{x^3},dx$. When you have a moment, I'd enjoy hearing your thoughts. I hope that you are staying healthy and safe my friend. – Mark Viola Jun 23 '20 at 19:36