Integration of an improper integral and the Cauchy principal value

I have been trying to evaluate the following integral: $$\int^{0}_{-\infty}e^{-i\omega t}dt$$ but I'm having trouble arriving at the correct result. My workings so far are as follows: $$\int^{0}_{-\infty}e^{-i\omega t}dt = \lim_{R\rightarrow\infty}\int^{0}_{-R}e^{-i\omega t}dt = \lim_{R\rightarrow\infty}\left(\int^{R}_{-R}e^{-i\omega t}dt - \int^{R}_{0}e^{-i\omega t}dt\right)\\ = 2\pi\delta(\omega) - \lim_{R\rightarrow\infty}\frac{i}{\omega}\left(e^{i\omega R}-1\right)\qquad\quad\quad\quad\;\;\;$$ but I'm stuck with how to proceed from here. (I know that the answer should be $$\int^{0}_{-\infty}e^{-i\omega t}dt=\pi\delta(\omega) +i\mathcal{P}\frac{1}{\omega}$$ where $\mathcal{P}$ denotes the Cauchy principal value).

Any tips on how to proceed would be much appreciated!

Solution 1:

The usual trick (search "Fourier transform of unit step" or Heaviside function) is to evaluate for $\epsilon>0$ : \begin{align} I_{\epsilon}(\omega)&:=\int^{0}_{-\infty}e^{(\epsilon-i\omega) t}dt\\ &=\frac 1{\epsilon-i\omega}\\ &=\frac {\epsilon+i\omega}{\epsilon^2+\omega^2}\\ &=\frac {\epsilon}{\epsilon^2+\omega^2}+i\frac {\omega}{\epsilon^2+\omega^2}\\ \end{align} As $\;\epsilon\to 0\,$ we obtain at the limit :

• $\dfrac {\epsilon}{\epsilon^2+\omega^2}\to \pi\,\delta(\omega),\quad$ since $\;\displaystyle \int_{-\infty}^{+\infty} \dfrac {\epsilon}{\epsilon^2+\omega^2}\,d\omega=\left.\arctan\dfrac {\omega}{\epsilon}\right|_{-\infty}^{+\infty}=\pi$
  (the Dirac delta distribution may be defined as the limit of the Lorentzian $\;\displaystyle\delta(\omega):=\lim_{\epsilon\to 0}\;\frac 1{\pi}\frac {\epsilon}{\epsilon^2+\omega^2}$)
• $\;\dfrac {\omega}{\epsilon^2+\omega^2}\to \mathcal{P}\dfrac 1{\omega},\quad$ (the singular part at $0$ is removed in a symmetrical way)