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

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