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I am trying to determine the Fourier transform of the Heaviside function $H$ as a tempered distribution. Consequently, for $\phi\in\mathcal{S}(\mathbb{R})$, I get

$$\begin{aligned} (\mathcal{F}H)(\phi)&=H(\mathcal{F}\phi) \\ &=\int_0^\infty(\mathcal{F}\phi)(\xi)\,d\xi \\ &=\int_0^\infty\left(\int_{-\infty}^\infty\phi(x)e^{-i\xi x}\,dx\right)\,d\xi \\ &=\int_{-\infty}^{\infty}\left(\int_0^\infty e^{-i\xi x}\,d\xi\right)\phi(x)\,dx\qquad\text{by Fubini} \\ &=\lim_{N\to\infty}\int_{-N}^N\left(\frac{i}{x} e^{-i\xi x}\bigg|_{\xi=0}^{\infty}\right)\phi(x)\,dx \\ &=\lim_{N\to\infty}\int_{-N}^N\left(\frac{i}{x}\lim_{R\to\infty}e^{-i\xi x}\bigg|_{\xi=0}^R\right)\phi(x)\,dx \\ &=i\lim_{N\to\infty}\int_{-N}^N\frac{\phi(x)}{x}\lim_{R\to\infty}e^{-iRx}\,dx-i\lim_{N\to\infty}\int_{-N}^N\frac{\phi(x)}{x}\,dx, \end{aligned}$$ so where am I going wrong? If I had $\epsilon\downarrow 0$ instead of $N\to\infty$, then this would certainly look closer to what I want, but I cannot see the mistake.

Jason Born
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  • @user1952009 This is wrong. The Fourier transform has a continuous linear extension to a continuous linear map $\mathcal{F}:\mathcal{S}'(\mathbb{R})\to\mathcal{S}'(\mathbb{R})$ and sonequently, for all $u\in\mathcal{S}'(\mathbb{R})$ and $\phi\in\mathcal{S}(\mathbb{R})$, one has $(\mathcal{F}u)(\phi)=u(\mathcal{F}\phi)$. – Jason Born Mar 26 '17 at 15:04
  • It depends on the definition you are using for the pairing, yes it isn't unitary you need a constant, and a complex conjugate – reuns Mar 26 '17 at 15:10
  • Also $\int_{-N}^N\frac{\phi(x)}{x} dx$ diverges whenever $\phi(0) \ne 0$. You can regularize it by looking at $\int_{-N}^N\frac{\phi(x)-\phi(0) e^{-a x^2}}{x} dx$ – reuns Mar 26 '17 at 15:16
  • The interchange of integration via Fubini (as in your question) is not legitimate. Instead, write $\lim_{N\to\infty}\int_{-\infty}^\infty \phi(x) \int_0^N e^{-i\xi x},d\xi,dx$, which is permitted. See THIS ANSWER for a rigorous development that follows the way you were headed. – Mark Viola Apr 22 '21 at 02:39

1 Answers1

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$$\int_0^\infty\left(\int_{-\infty}^\infty\phi(x)e^{-i\xi x}\,dx\right)\,d\xi =\int_{-\infty}^{\infty}\left(\int_0^\infty e^{-i\xi x}\,d\xi\right)\phi(x)\,dx$$ is not correct since $\int_0^\infty e^{-i\xi x}\,d\xi$ diverges for $\xi \in \mathbb{R}$.

You can regularize it with $\xi \in \mathbb{C},Im(\xi) < 0$ which is the same as saying that as $a \to 0$ : $$e^{-a^2 x} 1_{x > 0} \to 1_{x > 0}\qquad \text{in the sense of distributions}$$

So that $$( \mathcal{F} 1_{x > 0})(\phi) = \lim_{a \to 0} ( \mathcal{F} e^{-a^2 x} 1_{x > 0})(\phi)=\lim_{a \to 0}\int_{-\infty}^\infty \phi(x) \int_0^\infty e^{-a^2 t} e^{-itx}dtdx $$ $$= \lim_{a \to 0}\int_{-\infty}^\infty \frac{\phi(x)}{a^2+ix}dx =\lim_{a \to 0}\int_{-\infty}^\infty \frac{\phi(x)-\phi(0)1_{|x| < 1}}{a^2+ix}dx+\int_{-1}^1 \frac{\phi(0)}{a^2+ix}dx$$ $$= \int_{-\infty}^\infty\frac{\phi(x)-\phi(0)1_{|x| < 1}}{ix}dx+\lim_{a \to 0}\frac{\phi(0)\log(a^2+ix)}{i}|_{-1}^1$$ $$ = \lim_{\epsilon \to 0} \int_{|x| > |\epsilon|} \frac{\phi(x)}{ix}dx+ \pi \phi(0)$$

and hence $$\mathcal{F} 1_{x > 0} = p.v.(\frac{1}{i x})+\pi \delta(x)$$

reuns
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  • What is $\omega$? I guess you meant $\operatorname{p.v.}\left(\frac{1}{ix}\right)$. And isn't there normally a Dirac distribution and a few other things in the Fourier transform of $1_{x>0}$? – Jason Born Mar 26 '17 at 15:31
  • @JasonBorn do you like it now ? – reuns Mar 26 '17 at 15:51
  • With the convention the OP is using, you might need a complex conjugate so that $\mathcal{F} 1_{x > 0} = p.v.(\frac{1}{-i x})+\pi \delta(x)$ – reuns Mar 26 '17 at 15:52
  • Why does $\int_{-\infty}^{+\infty}\frac{\phi(x)-\phi(0)1_{|x|<1}}{ix} dx = \lim_{\epsilon\rightarrow 0}\int_{|x|>|\epsilon|} \frac{\phi(x)}{ix} dx$ ? – Jagerber48 Mar 27 '17 at 16:51
  • @Jagerber48 $\int_{-\infty}^{+\infty}\frac{\phi(x)-\phi(0)1_{|x|<1}}{ix} dx = \lim_{\epsilon \to 0}\int_{|x| > \epsilon}\frac{\phi(x)-\phi(0)1_{|x|<1}}{ix} dx =\lim_{\epsilon \to 0}\int_{|x| > \epsilon}\frac{\phi(x)}{ix} dx$ – reuns Mar 27 '17 at 17:43