Consider the function $$\phi (x) = \frac{1}{2 \pi i x} \{\text{exp}(2 \pi i x-1)$$
I know that the Fourier transform is \begin{align}\hat{\phi}(\omega) = \begin{cases} \frac{1}{2 \pi}, \ \ \ 0 \leq \omega <2\pi\\\ 0, \ \ \ \text{otherwise}\end{cases} \end{align}
Why $\hat{\phi}(\omega)$ is called one sided Fourier transform?
Best Answer
Because the typical kind of the Fourier transform is: $$ \begin{align}f(\omega)=\int_{-\infty}^\infty e^{i \omega t}f(t) dt\end{align}. $$ For your concrete case (one sided Fourier transform) is: $$ \begin{align}f(\omega)=\int_{0}^\infty e^{i \omega t}f(t) dt\end{align}. $$