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!
Best Answer
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 :
(the Dirac delta distribution may be defined as the limit of the Lorentzian $\;\displaystyle\delta(\omega):=\lim_{\epsilon\to 0}\;\frac 1{\pi}\frac {\epsilon}{\epsilon^2+\omega^2}$)