# Does the Fourier Transform exist for f(t) = 1/t?

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.

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

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