Fourier transform for $H(t)$ and test function

Question

Let $$H(t) = \begin{cases} 1 & t\gt0 \\ 0 & t\lt 0\end{cases}$$ I'm trying to find Fourier transform of $H(t).$ So we have $$\mathcal{F}\{H(t)\} = \int_{-\infty}^{+\infty}H(t)e^{-j\omega t}dt = \int_{0}^{+\infty}e^{-j\omega t}dt$$ Obviously this integral doesn't converge. So there are some ways to make this integral meaningful like introducing a damping factor. According to what I know from the distribution theory, If we want to view that as a distribution, we should see what happens when it acts on a test function $\phi(\omega)$. Then $$I =\int_{-\infty}^{+\infty}H(\omega)\phi(\omega)d\omega = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}H(t)e^{-j\omega t}dt\phi(\omega)d\omega = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}H(t)\phi(\omega)e^{-j\omega t}dtd\omega$$ Assuming changing order of integration is valid $$I = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}H(t)\phi(\omega)e^{-j\omega t}d\omega dt= \int_{-\infty}^{+\infty}H(t)\int_{-\infty}^{+\infty}\phi(\omega)e^{-j\omega t}d\omega dt$$ And I'm stuck here. The answer should be $$H(\omega) = \frac{1}{j\omega} + \pi\delta(\omega)$$I've seen many other pages on MSE related to this problem but didn't found an answer which continues my approach. Maybe I'm totally wrong?

Also in the distributional sense, it has been proved that $$H(\omega) = \pi \delta(\omega) + \mathrm{P} \frac{1}{j\omega}$$ So we have $\mathrm{P} \frac{1}{j\omega} = \frac{1}{j\omega}$?

Answer 1

I will denote the Fourier transform of $f$ by $\widehat{f}$, and I recall that by definition $$ \langle P(\tfrac{1}{x}),\varphi\rangle = \lim_{\varepsilon\to 0} \int_{|x|>\varepsilon}\frac{\varphi(x)}{x}\,\mathrm{d} x $$

So as you write, for any test function $\varphi\in C^\infty_c(\mathbb{R})$ we have $$ \begin{align*} \langle\widehat{H},\varphi\rangle &= \int_{\mathbb{R}} H(x)\,\widehat{\varphi}(x)\,\mathrm{d} x \\ &= \int_0^\infty\int_{\mathbb{R}} e^{-ixy}\varphi(y)\,\mathrm{d} y\,\mathrm{d}x \\ &= \lim_{n\to\infty} \int_0^n\int_{\mathbb{R}} e^{-ixy}\varphi(y)\,\mathrm{d} y\,\mathrm{d}x \end{align*} $$ Since $\varphi$ is compactly supported, there exists $a>0$ such that $\varphi=0$ out of $[-a,a]$, and since $\varphi$ and bounded, $$\int_0^n\int_{\mathbb{R}} |\varphi(y)|\,\mathrm{d} y\,\mathrm{d}x = \int_0^n\int_{-a}^a |\varphi(y)|\,\mathrm{d} y\,\mathrm{d}x < \infty $$ so we can use Fubini Theorem to get $$ \begin{align*} \langle\widehat{H},\varphi\rangle &= \lim_{n\to\infty} \int_{\mathbb{R}} \varphi(y) \int_0^n e^{-ixy}\,\mathrm{d}x\,\mathrm{d} y \\ &= \lim_{\varepsilon\to 0,\,n\to\infty} \int_{\varepsilon<|y|} \varphi(y)\, \frac{1-e^{-iny}}{i\,y}\,\mathrm{d} y \\ &= \langle P(\tfrac{1}{ix}),\varphi\rangle + \lim_{\varepsilon\to 0,\,n\to\infty} \int_{\varepsilon<|y|<a} \varphi(y)\, \frac{-e^{-iny}}{i\,y}\,\mathrm{d} y \end{align*} $$ Now remark that by doing the change of variable $y \to -y$ in the second integral when $y<0$, we have $$ \begin{align*} \int_{\varepsilon<|y|<a} \varphi(y)\, \frac{-e^{-iny}}{i\,y}\,\mathrm{d} y &= \int_{\varepsilon<y<a} \frac{\varphi(-y)e^{iny}-\varphi(y)e^{-iny}}{i\,y}\,\mathrm{d} y \\ &= \int_{\varepsilon<y<a} \frac{(\varphi(-y)-\varphi(0))\,e^{iny}-(\varphi(y)-\varphi(0))e^{-iny}}{i\,y}\,\mathrm{d} y \\ &\qquad + \varphi(0)\int_{\varepsilon<y<a} \frac{\,e^{iny}-e^{-iny}}{i\,y}\,\mathrm{d} y \\ &= \int_{\varepsilon<|y|<a} \psi(y)e^{-iny}\,\mathrm{d} y + \varphi(0)\int_{\varepsilon<y<a} \frac{2\sin(ny)}{y}\,\mathrm{d} y \end{align*} $$ where $\psi(y) = \frac{\varphi(0)-\varphi(y)}{iy}$ is a smooth function. To conclude remark that with $u = ny$ we get $$ \int_{\varepsilon<y<a} \frac{2\sin(ny)}{y}\,\mathrm{d} y = \int_{n\varepsilon<u<na} \frac{2\sin(u)}{u}\,\mathrm{d} u \underset{\varepsilon\to 0,\,n\to\infty}{\longrightarrow} \int_{0}^\infty \frac{2\sin(u)}{u}\,\mathrm{d} u = \pi $$ where one should first take the limit in $\varepsilon$ and then in $n$, while by Riemann-Lebesgue Lemma $$ \lim_{\varepsilon \to 0} \int_{\varepsilon<|y|<a} \psi(y)e^{-iny}\,\mathrm{d} y = \int_{|y|<a} \psi(y)e^{-iny}\,\mathrm{d} y\underset{n\to\infty}{\longrightarrow} 0 $$

Therefore $$\boxed{\langle\widehat{H},\varphi\rangle = \langle P(\tfrac{1}{ix}),\varphi\rangle + \pi\,\varphi(0)}$$ or equivalently, in the sense of distributions, we have the equality $$ \widehat{H} = P(\tfrac{1}{ix}) + \pi\,\delta_0 $$

Answer 2

Your test-function approach is correct as your guess, maybe with some constant factor distinct due to the $2\pi$ factor on the corresponding Fourier Transform definition you are using.

A year or two ago I was dealing with the same question and found The Fourier transform of the Heaviside function explaniation, not bad.

I hope this link still helps!

Answer 3

$\def\sign{\operatorname{sign}}$ Let's start with the $\sign$ function: $$ \sign(x) = \begin{cases} -1, & (x<0) \\ 1, & (x>0) \\ \end{cases} $$ Note that $\sign' = 2\delta$ and that $\sign$ is odd.

Taking the Fourier transform of $\sign' = 2\delta$ gives $i\xi \, \widehat{\sign}(\xi) = 2.$ Therefore $\widehat{\sign}(\xi) = \frac{2}{i\xi} + C\delta(\xi),$ where $\frac{1}{\xi}$ is the principal value distribution. Since $\sign$ is odd, so must be $\widehat{\sign}$. Therefore $C=0.$ Thus, $\widehat{\sign}(\xi) = \frac{2}{i\xi}.$

Now, $H(x) = \frac12(1+\sign(x))$ so $$ \hat{H}(\xi) = \frac12(\hat{1}(\xi) + \widehat{\sign}(\xi)) = \frac12(2\pi\,\delta(\xi) + \frac{2}{i\xi}) = \pi\,\delta(\xi) + \frac{1}{i\xi}. $$

Answer 4

As $r\to 0^+$, $$e^{-rt}1_{t >0}\to 1_{t>0}$$ in the sense of tempered distributions thus taking the Fourier of both side $$\frac1{r+i\omega}\to \widehat{1_{t >0}}$$ $\frac1{r+i\omega}$ is the distributional derivative of $-i\log(\omega-ir)$ and in $L^1_{loc}$ thus in the sense of distributions $$-i\log(\omega-ir)\to -i\log |\omega|-\pi 1_{\omega <0}$$

Thus, taking the distributional derivative of both side $$\frac1{r+i\omega}\to pv(\frac1{i\omega})+\pi\delta(\omega)$$

ie. $$\widehat{1_{t >0}}=pv(\frac1{i\omega})+\pi\delta(\omega)$$