# Fourier Transform of Heaviside Step Function

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.

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