2
$\begingroup$

I am hoping to calculate the Fourier transform of the sgn function directly from the definition of the Fourier Transform of a distribution. Most of the derivations I have seen calculate it by noticing the derivative of the sgn function or Heaviside step function is related to a dirac delta and working from there. I don't want to take this approach.

$$ s(t) = \text{sgn}(t) = \begin{cases} 1 & \text{for } t>0\\ 0 & \text{for } t=0\\ -1 & \text{for } t<0 \end{cases} $$

We can consider the corresponding distribution $s[f]$ defined by

\begin{align} s[f] = \int_{t=-\infty}^{+\infty} s(t) f(t) dt \end{align}

We then have, from the definition of the Fourier transform of a distrubution:

\begin{align} \tilde{s}[f] = s[\tilde{f}] \end{align}

Where $\tilde{f}$ is the Fourier Transform of $f$, $\tilde{f}(\omega) = \int_{t=\infty}^{+\infty} e^{i\omega t}f(t) dt$ From the approaches mentioned earlier I expect

\begin{align} \tilde{s}[f] = 2i P\int_{t= -\infty}^{+\infty} \frac{f(t)}{t} dt = 2i \lim_{\epsilon\rightarrow 0} \left( \int_{t= -\infty}^{\epsilon} \frac{f(t)}{t} dt + \int_{t= +\epsilon}^{+\infty} \frac{f(t)}{t} dt\right) \end{align}

Here is my approach which I can't get to give this result.

\begin{align} s[\tilde{f}] = \int_{\omega = -\infty}^{+\infty} s(\omega)\tilde{f}(\omega) d\omega =\\ =\int_{\omega = 0}^{+\infty} \tilde{f}(\omega) d\omega - \int_{\omega = -\infty}^{0} \tilde{f}(\omega) d\omega\\ \end{align}

We then perform a change of variables $\omega \rightarrow -\omega$ and swap the bounds on the integration to find

\begin{align} = \int_{\omega = 0}^{+\infty} \tilde{f}(\omega) - \tilde{f}(-\omega) d\omega \end{align}

\begin{align} \int_{\omega = 0}^{+\infty} \int_{t=-\infty}^{+\infty} (e^{i\omega t} - e^{-i\omega t}) f(t) dt d\omega = 2i\int_{\omega = 0}^{+\infty} \int_{t=-\infty}^{+\infty} \sin(\omega t) f(t)dtd\omega \end{align}

This is about where I get stuck. What I've tried so far is replacing the bounds on the integrals with variables and taking the limit as those variables go to infinity.

\begin{align} 2i\int_{\omega = 0}^{+\infty} \int_{t=-\infty}^{+\infty} \sin(\omega t) f(t)dtd\omega = 2i \lim_{a,b\rightarrow \infty} \int_{\omega = 0}^{b} \int_{t=-a}^{+a} \sin(\omega t) f(t)dtd\omega\\ = 2i \lim_{a,b\rightarrow \infty} \int_{t=-a}^{+a} \frac{1 - \cos(bt)}{t} f(t) dt \end{align}

At this point if I ignored the cosine term and also the fact that the taking the principal value is necessary I would have the correct answer but I can't see the justification for those moves.

I'll note that I work in physics, not math, so I didn't worry about whether the order of taking the limits or integration or anything matters. I assume it doesn't since $f(t)$ is a nice function which decays quickly enough since it is in the domain of the distribution. Maybe this is what I'm missing.

Is this approach tenable and if not why?

$\endgroup$
2
$\begingroup$

I have found the problem. As I hinted at in the question it has to do with the following.

$$ 2i \int_{\omega=0}^{+\infty}\int_{t=-\infty}^{+\infty} \sin(\omega t) f(t) dt d\omega = 2i \int_{t=-\infty}^{+\infty} \int_{\omega=0}^{+\infty} \sin(\omega t) f(t) d\omega dt $$

The swapping of the order of integration is not valid since the integrals in question do not converge on their own. The main message here is that direct computation of the integral is not going to be possible in this way. However, that doesn't mean we can't play tricks to compute the integral.

We step back to

$$ s[\tilde{f}] = \int_{\omega=0}^{+\infty} \int_{t=-\infty}^{+\infty}(e^{i\omega t} -e^{-i\omega t})f(t) dt d\omega $$

This integral is what we need to calculate. We note that this integral should converge since $f(t)$ is a 'nice' function. This means we can replace the integrand by

$$ (e^{i\omega t - a\omega}-e^{i\omega t - a\omega})f(t) $$

and take the limit as $a \to 0^+$. Noting $a>0$. By the dominated convergence theorem we can write.

$$ s[\tilde{f}] = \lim_{a \to 0^+} \int_{\omega=0}^{+\infty} \int_{t=-\infty}^{+\infty}(e^{i\omega t-a\omega} -e^{-i\omega t-a\omega})f(t) dt d\omega $$

The dominated convergence theorem allowed us to move the limit outside the integral. We now don't have convergence problems inside the integral so we can apply Fubini's theorem and determine:

$$ =\lim_{a \to 0^+} \int_{t=-\infty}^{+\infty} \left(\frac{-1}{it - a} - \frac{-1}{-it -a}\right)f(t) dt $$ $$ =2i\lim_{a \to 0^+} \int_{t=-\infty}^{+\infty} \left(\frac{t}{a^2+t^2}\right)f(t) dt $$

and it is well known that this integral gives the Caucy principal value as desired. For my and others reference I'll work out the details here.

$$ \lim_{a \to 0^+} \int_{t=-\infty}^{+\infty} \left(\frac{t}{a^2+t^2}\right)f(t) dt = \lim_{a \to 0^+} \lim_{\delta \to 0^+} \left( \int_{t = -\infty}^{-\delta}\left(\frac{t}{a^2+t^2}\right)f(t) dt + \int_{t = \delta}^{+\infty}\left(\frac{t}{a^2+t^2}\right)f(t) dt + \int_{t = -\delta}^{+\delta}\left(\frac{t}{a^2+t^2}\right)f(t) dt\right) $$

This follows because, so long as $a$ is non-zero we can split up the integral in the usual way. In the first two terms we can pass the $a$ limit through the $\delta$ limit and the integration because nothing singular is happening. For the final term we can perform the $\delta$ integration first and see that that term goes to zero. Putting it together

$$ \lim_{a \to 0^+} \int_{t=-\infty}^{+\infty} \left(\frac{t}{a^2+t^2}\right)f(t) dt = \lim_{\delta \to 0^+} \left( \int_{t = -\infty}^{-\delta}\frac{f(t)}{t} dt + \int_{t = \delta}^{+\infty}\frac{f(t)}{t} dt \right) = P\int_{t=-\infty}^{+\infty} \frac{f(t)}{t} dt $$

We then summarize the main result

$$ \tilde{s}[f] = s[\tilde{f}] = 2i P\int_{t=-\infty}^{+\infty} \frac{f(t)}{t} dt $$

which we can abbreviate as

$$ \tilde{s}(\omega) = 2iP\left(\frac{1}{\omega}\right) $$

edit: This answer was motivated by user1952009's response in Principal value not appearing in Fourier transform of the Heaviside function

$\endgroup$
1
$\begingroup$

After reading the question I have tried to find a derivation using distribution theory myself , but I was also surprised that I did not find any. Maybe you will find these lecture notes of interest, since there are several other derivations of FT pairs using tempered distributions, although I don't think $sgn$ in particular is derived :

The Fourier Transform and its Applications

One approach may be to consider $sgn(x) = 2 H(x) - 1$ where $H(x)$ is the Heaviside step function. Using the notion of differentiation in distribution theory, we can compute $\frac{dsgn(x)}{dx} = 2\frac{dH(x)}{dx} = 2\delta(x)$

Noting that $\hat f(\delta)=1$ and so $\hat f(2\delta)=2$, using the fact that $\hat f´(\omega)$ is $i\omega \hat f(\omega)$ (or $2\pi i$, depending on the definition) we have that $\hat f(sgn(w))$ = $\frac{2}{i\omega}$

$\endgroup$

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.