I strongly suspect, but am unable to prove, that the following limit holds (in the sense of distributions):
$$\lim_{t\to \infty} \frac{\exp(i \omega t) -1}{i\omega} = \pi \delta(\omega)+i\mathcal{P}\frac{1}{\omega}\tag{1}\label{eq}$$
where $\mathcal{P}$ is a Cauchy principal value operator. The first term is easy to see, since the real part $\sin(\omega t)/\omega$ is a nascent delta function. Note that this expression is very similar to the Sokhotski-Plemelj formula
$$\lim_{\epsilon\to 0}\frac{1}{\omega\pm i \epsilon}=\mp i\pi \delta(\omega)+\mathcal{P}\frac1\omega, $$
but this latter limit is easier to show.
How does one prove Eq. \ref{eq}, in particular the principal value term? The real reason I'm wondering is that I need to evaluate some more complicated limits, including
$$ \lim_{t\to \infty} \frac{\left(\exp(i \omega_1 t) -1\right)\left(\exp(-i\omega_2 t)-1\right)}{\omega_1 \omega_2}. $$
Here, the interesting case is $\omega_1=\pm \omega_2$. I originally believed
$$ \lim_{t\to \infty} \frac{\left(\exp(i \omega_1 t) -1\right)\left(\exp(-i\omega_2 t)-1\right)}{\omega_1 \omega_2} =\begin{cases}
\pi^2 \delta(\omega_1) \delta(\omega_2), \, \omega_1 \ne \omega_2 \\
2 \pi t \delta(\omega_1), \, \omega_1 = \omega_2.
\end{cases}, $$
but now I am convinced I am missing some principal value terms.
Best Answer
As a distribution, $\cos\omega t \to 0$ when $t \to \infty$ so $$ \omega \frac{\cos\omega t - 1}{\omega} = \cos\omega t - 1 \to 0 - 1 = -1 . $$
From this we can conclude that $$ \frac{\cos\omega t - 1}{\omega} \to -\mathcal{P}\frac{1}{\omega} + C \, \delta(\omega) $$ for some constant $C.$
But the left hand side is odd, so the right hand side must also be odd, which requires $C=0.$
Thus, $$ \frac{\cos\omega t - 1}{\omega} \to -\mathcal{P}\frac{1}{\omega} $$ and $$ \frac{\exp(i\omega t)-1}{i\omega} = -i\frac{\cos\omega t - 1}{\omega} + \frac{\sin\omega t}{\omega} \to i \, \mathcal{P}\frac{1}{\omega} + \pi\,\delta(\omega) . $$