### Author Topic: HA10 Problem 4  (Read 1813 times)

#### Victor Ivrii

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##### HA10 Problem 4
« on: March 26, 2015, 03:11:05 PM »
Find Fourier transform of $\theta(x)$ (Heaviside function).

#### Chaojie Li

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##### Re: HA10 Problem 4
« Reply #1 on: April 02, 2015, 08:32:20 PM »

#### Mark Nunez

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##### Re: HA10 Problem 4
« Reply #2 on: April 02, 2015, 09:02:43 PM »
A different derivation.

#### Victor Ivrii

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##### Re: HA10 Problem 4
« Reply #3 on: April 03, 2015, 12:33:49 PM »
Mark is completely correct; Chaojie needs to explain what is $\frac{1}{k}$ in the sense of distributions. If $\theta_-(x):=\theta (-x)=1-\theta(x)$ then
$\newcommand{\sgn}{\operatorname{sgn}} \sgn(x)=\theta(x)-\theta(-x)$
\begin{align}
&\hat{\theta}(k)= \frac{1}{2i\pi}(k-i0)^{-1}=\lim_{\varepsilon\to +0} \frac{1}{2i\pi}(k-i\varepsilon)^{-1},\\
&\hat{\theta_-}(k)= -\frac{1}{2i\pi}(k+i0)^{-1}=\lim_{\varepsilon\to +0} -\frac{1}{2i\pi}(k+i\varepsilon)^{-1},\\
&\hat{1}=\frac{1}{2i\pi} \bigl[ (k-i0)^{-1}-(k+i0)^{-1}\bigr]= \delta(k),\\
&\widehat{\sgn}(k)=\frac{1}{2i\pi} \bigl[ (k-i0)^{-1}+(k+i0)^{-1}\bigr] = \frac{1}{i\pi}pv (k^{-1})
\end{align}
with
$$pv(k^{-1}) (\varphi)=pv \int_{-\infty}^\infty k^{-1}\varphi(k)\,dk.$$