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I would like to compute the Fourier transform of the Heaviside function. To do this, I want to use the fact that : $$ F(vp \frac{1}{x}) = -2i\pi H+i\pi $$ where $F$ is the Fourier transform operator and $H$ is the Heaviside step function.

I know that this expression is correct but my problem is that I don't know how to find the correct expression of $F(H)$ from there.

Here is what I get : $$2i\pi H = F(vp\frac{1}{x})+i\pi \Leftrightarrow F(H)=-\frac{1}{2i\pi}F\bigg(F(vp\frac{1}{x})\bigg)+\frac{1}{2}\delta$$ where delta is the dirac distribution.

I know the correct final expression is: $$F(H)=\frac{1}{2i\pi}vp\frac{1}{x}+\frac{1}{2}\delta$$

But I don't really see how to obtain this final expression.

Arctic Char
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Peter
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1 Answers1

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Take the Fourier transform of both sides of $F(vp \frac{1}{x}) = -2i\pi H+i\pi.$

Then use the fact that $F(F(f(x))) = [2\pi]f(-x)$ (whether the factor $2\pi$ is there or not depends on your exact definition of the Fourier transform).

md2perpe
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  • Thank you for you answer, I was expecting something like this but is the following defintion always true ?$$F(F(f(x)))=[2\pi]f(-x)$$ Can you explain (or point to a good reference) where it comes from ? – Peter Aug 24 '21 at 14:13
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    It's a generally valid theorem. On Wikipedia it is written as $\mathcal{F}^2=\mathcal{P},$ where $\mathcal{P}$ is defined by $(\mathcal{P}f)(x)=f(-x).$ The Fourier inversion theorem is derived from this, and this can be derived from the Fourier inversion theorem. – md2perpe Aug 24 '21 at 15:56