2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ $\displaystyle\mathrm{H}\left(x\right) = \int_{-\infty}^{\infty}{\mathrm{e}^{\mathrm{i}kx} \over k - \mathrm{i}0^{+}}\,{\mathrm{d}k \over 2\pi\mathrm{i}}$ where $\displaystyle x \in \mathbb{R}\setminus\left\{0\right\}$ $\endgroup$ – Felix Marin May 3 '18 at 6:32
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.