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