I have a problem in evaluating the expression $$\int_{-\tau}^\infty {e^{i\omega t}dt}.$$ where the $\infty$ is taken in the distribution sense.
First, we have a well-known formula $$\int_0^\infty {e^{i\omega t}dt}=\pi\delta(\omega)+i\mathcal{P}\left(\frac{1}{\omega}\right),$$ which can be obtained by utilizing $$\int_0^\infty {e^{i\omega t}dt}=\lim_{\mu\rightarrow0^+}{\int_0^\infty{e^{-\mu t}e^{i\omega t}dt}},$$ where the limit of $\mu$ is taken w.r.t. the distribution topology.
I used the same strategy to compute the integration at the beginning, as follows.
$$\int_{-\tau}^\infty {e^{i\omega t}e^{-\mu t}dt}=\left.\frac{e^{(-\mu+i\omega)t}}{-\mu+i\omega}\right|_{-\tau}^\infty$$ $$=\frac{e^{(-\mu+i\omega)\tau}}{-\mu+i\omega}$$ $$=\frac{e^{-\mu\tau}e^{-i\omega\tau}(\mu+i\omega)}{\mu^2+\omega^2}.$$
Therefore, the integral $$\int_0^\infty {e^{i\omega t}dt}=\lim_{\mu\rightarrow0^+}{\int_0^\infty{e^{-\mu t}e^{i\omega t}dt}}$$ $$=\lim_{\mu\rightarrow 0^+}\frac{e^{i\omega\tau}e^{-\mu\tau}\mu}{\omega^2+\mu^2}+i\lim_{\mu\rightarrow 0^+}\frac{e^{i\omega\tau}e^{-\mu\tau}\omega}{\omega^2+\mu^2}.$$
This is where I got stuck. I know that $$\lim_{\mu\rightarrow 0^+}\frac{\mu}{\omega^2+\mu^2}=\pi\delta(\omega),$$ and that $$lim_{\mu\rightarrow 0^+}\frac{\omega}{\omega^2+\mu^2}=\mathcal{P}\left(\frac{1}{\omega}\right).$$ Indeed this is how we can obtain the classic formula for $\int_0^\infty {e^{i\omega t}dt}$. Here, how can I evaluate these two limits? How does the terms $e^{i\omega\tau}$ and $e^{-\mu\tau}$ affect the limit?
A naive guess is that since $e^{-\mu\tau}$ is very close to 1 when $\mu$ goes to 0, I can send $e^{-\mu\tau}$ to 1 first. But I can hardly justify this heuristic thinking.
Lastly, it would be best if anyone can provide any insight how could $\tau$ be pushed to infinity and recover another well known formula $$\int_{-\infty}^\infty {e^{i\omega t}dt}=2\pi\delta(\omega).$$
P.S., I can split the integral to two parts, and write $$\int_{-\tau}^\infty {e^{i\omega t}dt}=\left(\int_{-\tau}^0+\int_0^\infty\right)e^{i\omega t}dt,$$ and the second integral gives $\pi\delta(\omega)+i\mathcal{P}\left(\frac{1}{\omega}\right)$. However, the problem, in my mind, merely was shifted to the first integral, and is still there. Ironically, this first integral in the usual sense is trivial and the result is simply $$\int_{-\tau}^0{e^{i\omega t}dt}=\frac{1-e^{-i\omega\tau}}{i\omega}.$$ However, I don't know how to look at this expression in the viewpoint of distributions, not to mention pushing $\tau$ to infinity to get $\int_{-\infty}^\infty e^{i\omega t}dt$.
Thanks in advance for any suggestions,
Zheng
Use the translation formula: $$ \int_{-\tau}^\infty e^{i\omega t} \, dt = \{ s = t + \tau \} = \int_0^\infty e^{i \omega (s-\tau)} \, ds = e^{-i \omega \tau} \int_0^\infty e^{i \omega s} \, ds $$