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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!

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  • $\begingroup$ $\displaystyle\mathrm{H}\left(x\right) = \int_{-\infty}^{\infty}{\mathrm{e}^{\mathrm{i}kx} \over k - \mathrm{i}0^{+}}\,{\mathrm{d}k \over 2\pi\mathrm{i}}$ where $\displaystyle x \in \mathbb{R}\setminus\left\{0\right\}$ $\endgroup$
    – Felix Marin
    May 3, 2018 at 6:32

1 Answer 1

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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