I've been trying to compute the fourier transform of $\operatorname{sgn}(x)$, but I'm having trouble with the complex exponential at infinity. The issue is the following: by definition we have


If we apply the definition of $\operatorname{sgn}$, we can easily see that

$$(\mathcal{F}[\operatorname{sgn}(x)](k),\phi(k))=\dfrac{1}{2\pi}\int_{-\infty}^{\infty}\phi(k)\left(\int_{-\infty}^0-e^{ikx}dx+\int_{0}^\infty e^{ikx}dx\right)dk.$$

Now on the inner integrals, if we integrate directly we would have

$$\int_{-\infty}^0-e^{ikx}dx+\int_{0}^\infty e^{ikx}dx=\int_0^\infty e^{ikx}-e^{-ikx}dx,$$

but this is

$$\int_{-\infty}^0-e^{ikx}dx+\int_{0}^\infty e^{ikx}dx=2i\int_0^\infty \sin{kx}dx$$

which we know that doesn't exist.

In that sense something is clearly wrong here but I'm not being able to see what. I believe my method is totally wrong here, but I can't see why.

What am I doing wrong? And how can we compute this fourier transform correctly?

  • $\begingroup$ You can't compute it directly, at least not like that, as far as I know. You might find these examples instructive. $\endgroup$ – suneater Mar 31 '16 at 22:17

The problem with writing down an improper integral $\int_{-\infty}^\infty$ and then maybe trying to find $\lim_{L\to\infty} \int_{-L}^L$ is that this truncation is too rough for such heavy-tailed function as $\operatorname{sign}$. This is why you can't get anything to converge.

The link given by zahbaz describes a better approximation, $e^{-\epsilon |x|} \operatorname{sign}x$ where $\epsilon\to 0$. For these functions the transform can be computed directly: $$F_\epsilon(\omega) = -\frac{2i\omega}{\omega^2+\epsilon^2}$$ and the limit as $\epsilon\to 0$ exists: $2/(i\omega)$. Note that the Fourier transform is continuous on the space of distributions: distributional convergence $f_n\to f$ implies distributional convergence $\hat {f_n}\to \hat f$.

A less direct, but easier, way is to recall that the distributional derivative of $\operatorname{sign}x$ is $2\delta_0$, and the Fourier transform of $\delta_0$ is just the constant function $2$ (under your normalization of the transform). Since $f'$ transforms to $ i\omega \hat f(\omega)$, it follows that the Fourier transform of $\operatorname{sign} x$ is $2/(i\omega)$.

  • $\begingroup$ Thanks for the answer. The second method is very nice, but there's something I'm unsure. My issue is that in the problem I was working, which asks to compute this Fourier transform, in the answer it is said that the Fourier transform of $\operatorname{sgn}(x)$ is $2i \operatorname{Pv}\frac{1}{k}$. By this method we get $-2i/k$. Why is there such a difference? Thanks very much again! $\endgroup$ – Gold Mar 31 '16 at 23:26
  • $\begingroup$ Maybe they defined the transform as $\int f(x) e^{ik x}$? I don't know. Including Pv is the correct thing to do, since $1/k$ is not a distribution. $\endgroup$ – user147263 Mar 31 '16 at 23:30
  • $\begingroup$ Yes, the transform was defined as $\int f(x) e^{ikx}$. Now I've noticed this takes care of the sign of $i$ and thus is the reason for the difference. On the other hand, the $\operatorname{Pv}$ is something I still didn't get. I mean, we have that $\mathcal{F}[\operatorname{sgn}'(x)]=-ik\mathcal{F}[\operatorname{sgn}(x)]$ Thus since $\mathcal{F}[\operatorname{sgn}'(x)]=2$ we obviously have $\mathcal{F}[\operatorname{sgn}(x)]=2i/k$. The $\operatorname{Pv}$ didn't appear naturally there. It is something we include by hand then or does it appear naturally somehow and I'm missing it? $\endgroup$ – Gold Mar 31 '16 at 23:40
  • $\begingroup$ Manipulations with distributions are not as straightforward as with functions; it's not just something we do pointwise. We can't even multiply or divide distributions in general. Here, we look for distribution $T$ such that the product of $T$ with the function $-ik$ is $2$. Such a distribution happens to exist: it is $2i Pv (1/k)$. $\endgroup$ – user147263 Mar 31 '16 at 23:46
  • $\begingroup$ That said, many times people will simply write $1/k$ and omit Pv until someone presses them about the meaning of $1/k$ in this context. It's something that only becomes an issue when your calculations yield a function that is not locally integrable. $\endgroup$ – user147263 Mar 31 '16 at 23:47

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.