What is one sided Fourier transform?

Question

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?

Answer 1

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: $$ \begin{align}f(\omega)=\int_{0}^\infty e^{i \omega t}f(t) dt\end{align}. $$

Answer 2

I think there are perhaps some problems with both the original question and previously accepted answer.

Note that

$$\int\limits_0^{2\pi}\frac{1}{2\pi}e^{i x \omega}\,d\omega=\frac{e^{2 \pi i x}-1}{2 \pi i x}\tag{1}$$

so I believe $\phi(x)$ should be defined as

$$\phi(x)=\frac{e^{2 \pi i x}-1}{2 \pi i x}\tag{2}$$

where the $-1$ is external to the exponentiation operation.

Also note that

$$\mathcal{F}_x[\phi (x)](\omega )=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty} \phi(x)\,e^{-i w x}\,dx=\frac{\text{sgn}(2\pi-\omega)+\text{sgn}(\omega)}{4\pi}=\left\{\begin{array}{cc} \frac{1}{2 \pi } & 0<\omega <2 \pi \\ \frac{1}{4 \pi } & \omega =0\lor \omega =2 \pi \\ 0 & \text{Otherwise} \end{array}\right.\tag{3}$$

and

$$\mathcal{F}_{\omega}^{-1}\left[\frac{\text{sgn}(2 \pi -\omega)+\text{sgn}(\omega)}{4 \pi }\right](x)=\int\limits_{-\infty}^{\infty}\frac{(\text{sgn}(2 \pi-\omega)+\text{sgn}(\omega))}{4 \pi } e^{i x \omega}\,d\omega =\frac{e^{2 \pi i x}-1}{2 \pi i x}=\phi(x)\tag{4}$$

so it appears the Fourier transform of $\phi(x)$ is not one-sided as defined in the previously accepted answer and associated comments.