Is it possible using the following property of Fourier transforms;
$$\frac{d^nf(x)}{dx^n} = 2\pi i\nu \hat{f}(x)$$
to show that the Fourier transform of the heaviside step function is equal to;
$$\hat{f}[\theta(t)] = \frac{1}{2\pi i \nu}$$
Using the property listed above we can write:
$$\hat{f}[\theta'(t)] = 2\pi i\nu\hat{f}[\theta(t)]$$ But since we know the (distributional) derivative of the Heaviside step function is the Dirac delta function.
$$ \hat{f}[\delta(t)] = 2\pi i\nu\hat{f}[\theta(t)]$$
$$\implies \hat{f}[\theta(t)] = \frac{\hat{f}[\delta(t)]}{2\pi i \nu}$$
And since the Fourier transform of the delta function is just $1$ we get:
$$ \hat{f}[\theta(t)] = \frac{1}{2 \pi i \nu}$$
I am unsure if this process is correct or rigorous since my knowledge and understanding of distributions and generalised functions is limited.
If this method presented above is indeed allowed I am unsure why this answer contradicts the answer shown on Mathworld which states the Fourier transform of the Heaviside step function is equal to:
$$ \frac{1}{2}\bigl[\delta(k)\space – \frac{i}{\pi k}\bigr]$$
There is also another article here https://www.cs.uaf.edu/~bueler/M611heaviside.pdf which does not return my answer.
Any comments, corrections or answers are greatly appreciated.
Thanks!
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{H}\pars{x} & = \int_{-\infty}^{\infty}{\expo{\ic kx} \over k - \ic 0^{+}}\, {\dd k \over 2\pi\ic} \\[5mm] \hat{\mrm{H}}\pars{k} & = {1 \over k -\ic 0^{+}}\,{1 \over \ic} = -\ic\bracks{\mrm{P.V.}{1 \over k} + \ic\pi\,\delta\pars{k}} = -\ic\,\mrm{P.V.}{1 \over k} + \pi\,\delta\pars{k} \end{align}