Thread: How do I calculate this Fourier transform?

Hi!

How can I calculate the Fourier transform of the sgn(x)? Now, I know that the answer should be a factor of 1/(iv), v angular frequency, depending on which Fourier transform that is uses. But how can I show this?

Which version of the Fourier Transform are you using?

I'm using the non-unitary one.

So,

$\displaystyle \displaystyle \hat{f}(\nu)=\mathcal{F}[f(x)]=\int_{-\infty}^{\infty}f(x)\,e^{-i\nu x}\,dx?$

Exactly

So,

$\displaystyle \displaystyle\mathcal{F}[\text{sgn}(x)]=\int_{-\infty}^{0}(-1)e^{-i\nu x}\,dx+\int_{0}^{\infty}(1)e^{-i\nu x}\,dx,$ right? Any ideas from here?

Originally Posted by Ackbeet

So,

$\displaystyle \displaystyle\mathcal{F}[\text{sgn}(x)]=\int_{-\infty}^{0}(-1)e^{-i\nu x}\,dx+\int_{0}^{\infty}(1)e^{-i\nu x}\,dx,$ right? Any ideas from here?

Hm, well, the integrals do not converge, however, maybe it's possible to let the limits of the integrals approach infinity and see what curve you'll get:

$\displaystyle \displaystyle\mathcal{F}[\text{sgn}(x)]=\int_{-\infty}^{0}(-1)e^{-i\nu x}\,dx+\int_{0}^{\infty}(1)e^{-i\nu x}\,dx$

$\displaystyle =\displaystyle\lim_{a\rightarrow\infty}\ \int_{-a}^{0}(-1)e^{-i\nu x}\,dx+\int_{0}^{a}(1)e^{-i\nu x}\,dx $

$\displaystyle =\displaystyle\lim_{a\rightarrow\infty}\ (-1)\displaystyle{\frac{1}{iv}}\left(\displaystyle{1-e^{-i(-a)x}\right)+(+1)\displaystyle{\frac{1}{iv}}\left(\ displaystyle{e^{-iax}-1\right)$

$\displaystyle =\displaystyle\lim_{a\rightarrow\infty}\ \displaystyle{\frac{i}{v}}\left(2-e^{iax}-e^{-iax}\right)$

$\displaystyle =\displaystyle\lim_{a\rightarrow\infty}\ \displaystyle{\frac{2i}{v}}(1-\cos(ax))$

Then the cosine term should be possible to remove, but I don't know how to show that you can do it.

Yes, I think you do need to do the Cauchy principal value, which is what you're doing. I'm not sure you're doing it quite right. You're good up to here:

$\displaystyle =\displaystyle\lim_{a\rightarrow\infty}\left[ \int_{-a}^{0}(-1)e^{-i\nu x}\,dx+\int_{0}^{a}(1)e^{-i\nu x}\,dx\right]$

$\displaystyle =\displaystyle\lim_{a\rightarrow\infty}\left[ (-1)\frac{1}{-i\nu}\,e^{-i\nu x}\Big|_{-a}^{0}+\frac{1}{-i\nu}\,e^{-i\nu x}\Big|_{0}^{a}\right]$

$\displaystyle =\displaystyle\lim_{a\rightarrow\infty}\left[ \frac{1}{i\nu}\,e^{-i\nu x}\Big|_{-a}^{0}-\frac{1}{i\nu}\,e^{-i\nu x}\Big|_{0}^{a}\right]$

$\displaystyle =\displaystyle\frac{1}{i\nu}\lim_{a\rightarrow\inf ty}\left[ (1-e^{i\nu a})-(e^{-i\nu a}-1)\right]$

$\displaystyle =\displaystyle\frac{2}{i\nu}\lim_{a\rightarrow\inf ty}\left[1-\cos(a\nu)\right]$

I don't think that limit is going to exist. I've fiddled around with various ideas: change of variables in the first integral, for example, but it leads to the same limit. One thing I have not done is try contour integration. You might be able to get that to work. I am, unfortunately, fast approaching the limits of my knowledge. Wiki says the answer is $\displaystyle 2/(i\nu).$ But, like you, I'm not entirely sure how to get there.

Maybe CB or mr fantastic or Opalg could weigh in and help out.

Hehe, seems like I managed to screw it up in two ways when I performed the calculations

Maybe the answer lies in the way the Fourier transform is later used. While the integral itself does not converge (I don't think contour integration will work here; the integrand would then have to approach 0 as x goes to infinity), the scalar product of the transform and any given test function will converge. I don't know if this is enough to conclude that the cosine term can be removed though but I think it may be something like that.

Good thinking. That would certainly be in line with the distribution idea.