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.

  • 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


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 $$ \mathcal{F}\left\{\frac{1}{\pi t}\right\}=-i \operatorname{sgn}(\omega). $$

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

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