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.


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 $$ \frac{\exp(i\omega t)-1}{i\omega} = -i\frac{\cos\omega t - 1}{\omega} + \frac{\sin\omega t}{\omega} \to i \, \mathcal{P}\frac{1}{\omega} + \pi\,\delta(\omega) . $$

  • $\begingroup$ I realized that it practically boils down to $$ \frac{\cos\omega t - 1}{\omega} = \cos\omega t \frac{1}{\omega} - \frac{1}{\omega} \to 0 \cdot \frac{1}{\omega} - \frac{1}{\omega} = - \frac{1}{\omega}. $$ $\endgroup$
    – md2perpe
    Sep 8 '20 at 12:04
  • $\begingroup$ Can you explain in a little more detail/rigor where the principal value comes from? Would it be true that, say, $\frac{\cos^2(\omega t)}{\omega^2} = \frac{1+\cos(2\omega t))}{2\omega^2} \to {\cal P}\frac{1}{2\omega^2}$? $\endgroup$ Sep 10 '20 at 0:27
  • $\begingroup$ @mikefallopian. The principal value distribution $\mathcal{P}\frac{1}{\omega}$ is the almost unique distribution satisfying $\omega \, u(\omega) = 1.$ With "almost unique" I refer to that one can add multiples of $\delta(\omega).$ $\endgroup$
    – md2perpe
    Sep 10 '20 at 6:36
  • $\begingroup$ Up to scalar multiples, it is the unique odd distribution satisfying $\omega\cdot u=1$ on $\mathbb R$. $\endgroup$ Dec 24 '20 at 23:15
  • 1
    $\begingroup$ The principal-value integral against $1/x$ is the unique odd distribution $\omega$ satisfying $x\cdot \omega=1$. $\endgroup$ Dec 25 '20 at 16:16

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