# What is the one-sided Fourier transform of a constant?

A definition of the Fourier transform commonly used is (I always forget which convention of normalization to use) \begin{align}f(\omega)=\int_{-\infty}^\infty e^{i \omega t}f(t) dt\end{align} For a constant function $$f(t)=1$$, this evaluates to the Dirac delta function, \begin{align}\int_{-\infty}^\infty e^{i \omega t}dt = 2\pi\delta(\omega)\end{align} Question:

What is the one-sided Fourier transform of a constant function\begin{align}\int_{0}^\infty e^{i \omega t}dt = ?\end{align}

$$\int_0^\infty e^{i\omega t}e^{-\epsilon t}dt= \frac 1{i\omega -\epsilon}$$ So the Fourier transform of the Heaviside step function $$\theta(t)$$ is the $$\epsilon\to 0$$ limit $$\mathcal{F}[\theta](\omega)= \lim_{\epsilon\to 0} \left\{\frac{-i}{\omega-i\epsilon}\right\}= P\left(\frac{1}{i\omega}\right)+\pi \delta(\omega),$$ where "$$P$$" is the principal part distribution. (I added edit to show connection with @Thomas Fritsch's answer)
You look for the Fourier transform of the step function: $$f(t)=\begin{cases} 1, &\text{for }t > 0 \\ 0, &\text{for }t < 0 \end{cases}$$
$$F(\omega)=\frac{1}{i\omega} + \pi\delta(\omega)$$
Depending on your sign and normalization convention you may need to modify this by a minus sign or by a $$2\pi$$ factor.