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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

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

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?

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  • $\begingroup$ You can't compute it directly, at least not like that, as far as I know. You might find these examples instructive. $\endgroup$ – zahbaz Mar 31 '16 at 22:17
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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 $2/(i\omega)$.

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  • $\begingroup$ 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! $\endgroup$ – user1620696 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? $\endgroup$ – user1620696 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

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