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!