9
$\begingroup$

My professor says that the following function has a Fourier Transform:

$$f(t) = \frac{1}{\pi t}$$

He said that all I have to do is apply some of the Fourier Transform properties and not the direct integral definition of the Fourier Transform to find it:

However, My book for the class claims that no Fourier Transform exists for the function:

$$f(t) = \frac{1}{t}$$

So I'm guessing that there is none, since I can't seem to figure out what property to use to find the Fourier Transform of that function. The $\frac{1}{\pi}$ term doesn't matter since it is a constant and can come out of the integral. The closet property that seems to maybe yield a result is the "Duality Property". An example from my book asks to find the Fourier Transform of the following function:

$$ f(t) = \frac{10}{t^2} $$

which from a standard Fourier transform table and using the duality principle is easily seen as:

$$ \mathfrak{F}[f(t)] = -10\pi |\omega| $$

However, that function has $t^2$ term. I'm stuck, anybody have a way to tackle this.

$\endgroup$
4
  • 2
    $\begingroup$ $\mathfrak{F}_t\left [\frac1t\right ](\omega)=i\sqrt{\frac{\pi}{2}}\operatorname{sgn}(\omega)$ $\endgroup$
    – UserX
    Nov 22, 2014 at 17:02
  • $\begingroup$ What is that function UserX? I don't have it in my table of Fourier Transforms, is it the sinc(x) function? $\endgroup$ Nov 22, 2014 at 17:08
  • 3
    $\begingroup$ How about use that multiplication-by-polynomials transforms to differentiation, then write $1/t = t/t^2$, then Fourier transform that to $(d/dt) F(1/t^2)$? $\endgroup$
    – Neal
    Nov 22, 2014 at 17:08
  • 1
    $\begingroup$ @ Neal, wow I didn't think of that. That's legit. Thank you. $\endgroup$ Nov 22, 2014 at 17:10

1 Answer 1

13
$\begingroup$

Consider the signum function $\operatorname{sgn}(t)$ which is defined as $$ \operatorname{sgn}(t)=\begin{cases}1& t>0\\ -1 & t<0\end{cases} $$ Consider the odd two sided exponential function $f_{\alpha}(t)$ defined as $$ f_{\alpha}(t)=\begin{cases} \operatorname{e}^{-\alpha t}& t>0\\ -\operatorname{e}^{\alpha t} & t<0\end{cases} $$ where $\alpha>0$.

Defining the Fourier transform of $f(t)$ as $\mathcal{F}\{f(t)\}=\int_{-\infty}^{\infty}f(t) e^{-i\omega t}\operatorname{d}\!t$, we find that the Fourier transform of $f_{\alpha}(t)$ is $$ \mathcal{F}\{f_{\alpha}(t)\}=-\frac{1}{\alpha-i\omega}+\frac{1}{\alpha+i\omega}=-\frac{2i\omega}{\alpha^2+\omega^2} $$ .

As we let $\alpha\to 0$ the exponential function resembles more and more closely the signum function, i.e. $\lim_{\alpha\to 0}f_{\alpha}(t)=\operatorname{sgn}(t)$; so we have $$ \mathcal{F}\{\operatorname{sgn}(t)\}=\lim_{\alpha\to 0}\mathcal{F}\{f_{\alpha}(t)\}=\frac{2}{i\omega} $$ Finally, by the duality property, we find $$ \mathcal{F}\left\{\frac{1}{\pi t}\right\}=-i \operatorname{sgn}(\omega). $$

$\endgroup$
1
  • 1
    $\begingroup$ Why is the limit justified? $\endgroup$
    – LL 3.14
    Nov 19, 2021 at 17:09

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.