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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!

Will
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    Correct me if I'm wrong, but are you sure the integral has a value? The integrals of $\sin$ and $\cos$ do not converge. – Chris Sanders Jul 07 '16 at 10:31
  • I think that the $\delta$ on the right is a Dirac-delta, i.e., when $\omega$ is zero, you get an infinite value, and for $\omega$ nonzero, you get something involving a script $P$, which I do not recognize. In short: this isn't a question about a calculus-class type of integral, but probably something involving distribution-fu as used in signal processing. – John Hughes Jul 07 '16 at 10:36
  • @ChrisSanders According to a set of notes that I read, and a paper on the arXiv, it does. The result that I put in brackets in my original post is what they quote the value of the integral as being. – Will Jul 07 '16 at 10:39
  • @JohnHughes Yes, the $\delta$ is a Dirac-delta, and the $\mathcal{P}$ is their notation for the Cauchy principal value of an improper integral. I'm completely stumped as to how to arrive at the quoted result, but I have to admit, I'm not very familiar with the usage of Cauchy principal values. – Will Jul 07 '16 at 10:41
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    See This Answer. – Mark Viola Sep 08 '22 at 21:19

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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)
Raymond Manzoni
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  • Thanks very much, this is just what I was looking for! Is there any literature on defining the delta function as the limit put in your answer (I've never seen it defined that way before and I'm curious to learn more)?! Also, a few questions. 1. What happens to the term $e^{(\epsilon -i\omega)(-\infty)}$? I can see that $e^{-\epsilon\infty}=0$, but what about the imaginary part (is it simply that it is finite and so can be neglected)? 2. Why does $\frac{\omega}{\epsilon^{2}+\omega^{2}}\rightarrow\mathcal{P}\frac{1}{\omega}$ as $\epsilon\rightarrow 0$? ... – Will Jul 07 '16 at 12:14
  • ...The function $\frac{1}{\omega}$ doesn't appear to be ill-defined?! – Will Jul 07 '16 at 12:19
  • Let's start by the end : $\dfrac{1}{\omega}$ doesn't define a distribution but $\mathcal{P}\dfrac{1}{\omega}$ does! The point is that $\dfrac{1}{\omega}$ can't be used in an integral like $;\displaystyle \int_a^b \frac{f(x)}x,dx,$ with $a<0<b$ (and $f$ continuous with $f(0)\neq 0$) because of the singularity at $0$ but $;\displaystyle\mathcal{P} \int_a^b \frac{f(x)}x,dx=\lim_{\epsilon\to 0},\int_a^{-\epsilon} \frac{f(x)}x,dx+\int_{\epsilon}^b \frac{f(x)}x,dx$. – Raymond Manzoni Jul 07 '16 at 13:00
  • has a fine behaviour. Distributions may always be derived or integrated and thus you'll find things like $\mathcal{P}\dfrac{1}x$ or $Pf\dfrac{1}{x^2}$ ("pseudo-function") instead of the usual negative powers.

    Concerning $;\displaystyle e^{(\epsilon -i\omega)(-\infty)}$ take the absolute value and observe that $;\displaystyle \left|e^{(\epsilon -i\omega)(-R)}\right|=e^{-\epsilon R}$ ($|e^{ix}|=1$ !).

    An interesting (tragic ;-)) reference is this paper "The Fourier transform of the Heaviside function: a tragedy". Cheers,

     – Raymond Manzoni Jul 07 '16 at 13:10
  • Thanks for the details. What is the actual integral form of $\mathcal{P}\frac{1}{\omega}$ out of interest? – Will Jul 07 '16 at 13:11
  • Your question is a little unclear ($\mathcal{P}$ acts a little like a filter removing a symmetrical interval around $0$ and excluding the parts larger than $+R$ and smaller than $-R$). See for example here. – Raymond Manzoni Jul 07 '16 at 13:17
  • Ah ok. I think it's becoming a little clearer now. Thanks! – Will Jul 07 '16 at 13:29
  • Glad it clarified things @Will. Excellent continuation! – Raymond Manzoni Jul 07 '16 at 13:31
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    HERE is another way forward aligned with the OP's attempted solution. – Mark Viola Sep 08 '22 at 21:19