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Long story short: I'm looking for a good way of showing that the Fourier transform of $1/x$ is a sign function.


Motivation and why this has been a problem:

I'm dealing with an equation similar to the Burgers equation in the form:

$$u_t = \varepsilon u_{xx} - u\cdot u_x$$

with initial conditions

$$u(0,x)=g'(x)$$

where $u_x$ denotes $\frac{\partial u}{\partial x}$.

In order to better understand my problem, I was looking at different ways of dealing with the Burgers equation.

It is well known that we can write the solutions of this equation as

$$u(t,x)= -2\varepsilon\sum_{j=1}^{2N}\frac{1}{x-z_j(t)}$$

where $z_j(t)$ are complex poles of the solution. In order to ensure the solutions will be real-valued, the poles are chosen such that

$$z_j = z^*_{j+N}$$ for $j=1,\dots,N$.

The problem arises when we look for periodic solutions of this equation. The literature I found about it suggests looking at periodic solutions of period $2\pi$ written as:

$$u(t,x)= -2\varepsilon\sum_{k=-\infty}^{\infty}\sum_{j=1}^{2N}\frac{1}{x-z_j(t)+2\pi k}$$

and then using the Poisson summation formula to rewrite this as a cotangent. This is the problem, I'm struggling to find literature about this representation.

The Poisson formula is:

$$\sum_{k=-\infty}^{\infty}f(x+2\pi k)=\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{2\pi inx}$$

So the main problem is to define the Fourier transform of $\frac{1}{x}$, which all the literature indicates should be a sign function. The best hint I've found on the problem was to consider the function

$$G(x,\alpha)=\left\{\array{e^{-\alpha x}, \hspace{1cm} x>0 \\ -e^{\alpha x}, \hspace{1cm} x<0 }\right.$$

Compute the Fourier transform of the function $G$ and then take the limit for $\alpha\rightarrow 0$. This would show that the Fourier transform of the of the sign function is $\frac{1}{x}$ in a sense.

My problem with it is that I can't seem to justify that the transform of the limit is the limit of the transform in this case. Moreover, even if it is true, this calculation doesn't yield a good definition for $\hat{f}(0)=sign(0)$, which is needed for the Poisson formula.

I would have simply have given up on a strong definition of this transform if it weren't for the multiple sources claiming that it exists and equals the sign function (but lacking a thorough proof).

Does anyone know of a good reference where I could read about it?

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  • $\begingroup$ $1/x$ is not locally integrable and thus doesn't have a FT. What is true is that the FT of (the tempered distribution) $\textrm{PV}(1/x)$ (principal value) is $c\textrm{ sign}(x)$, but this is a standard fact and not a suitable question for this site. Please use math.stackexchange.com instead. $\endgroup$ Commented Sep 20, 2023 at 19:38
  • $\begingroup$ I know this is a standard fact. My point is that I have found countless times (such as sciencedirect.com/topics/mathematics/hilbert-transform or sciencedirect.com/topics/psychology/hilbert-transform) that the Fourier transform of $1/x$ in the usual sense is being used as the sign function. My question is exactly why that is and I would have settled for a Cauchy Principal Value kind of integral, but many authors seem to imply otherwise. $\endgroup$
    – Rafael
    Commented Sep 20, 2023 at 19:48
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    $\begingroup$ There's a few ways to prove the cotangent identity you are looking for that avoid Poisson summation, I think. Have a look at this link: math.stackexchange.com/questions/1849878/… $\endgroup$ Commented Sep 20, 2023 at 22:57
  • $\begingroup$ You are right. I didn't need to look at the Poisson formula. I guess I was fixated on that because of the Hilbert transform I was also looking at. In the end I started going back at some definitions after the comments and realized, my confusion came from the fact that the papers I was reading were not careful defining the Hilbert transform and omitted the fact, that the integrals were principal valued. $\endgroup$
    – Rafael
    Commented Sep 21, 2023 at 1:20

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