I am trying evaluate

$I(\omega) = \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^{\infty} dx \log(i(x-i\epsilon)) e^{i\omega x}$

The answer for $\omega \in \mathbb{R}$ should be a distribution.
My progress so far is evaluating $I$ for $\omega \neq 0$ using contour integration, which is straightforward: I choose the branch cut at negative real values of the log, the branch cut of the integrand is on the positive imaginary axis. For $\omega < 0$ I can close the contour in the LHP and get $I = 0$. If $\omega > 0$ I have to close the contour in the UHP and integrate around the branch cut.

Since the integral along the two quarter circles and the small semi circle vanishes I get

$I =\lim_{\epsilon \rightarrow 0^+} i \int_{\epsilon}^{\infty}d\xi~ e^{-\omega \xi} \Big{[} \log(\xi) + i\pi - (\log(\xi) - i \pi) \Big{]} = - 2\pi \frac{1}{\omega}$

However if $\omega = 0$ I cant use contour integration. In a paper I found the result:

$I= \lim_{\epsilon \rightarrow 0^+} i \Big{[} \frac{\log(-\omega - i\epsilon)}{-\omega - i\epsilon} - \frac{\log(-\omega + i\epsilon)}{-\omega + i\epsilon} \Big{]}$

but I do not know how to obtain this equality and it is also not explained there.

  • $\begingroup$ $- 2\pi \log(\omega) \delta(\omega) - 2\pi \Theta(\omega) \frac{1}{\omega}$ doesn't make any sense, you should not follow a paper mentioning it. $\endgroup$ – reuns Aug 11 at 11:36
  • $\begingroup$ @reuns Ok I deleted the expression. It is also not exactly whats written in the paper. Does the first equality make sense? $\endgroup$ – lomby Aug 11 at 11:54
  • $\begingroup$ No, your integral doesn't converge so it doesn't make sense to use contour integration. $\endgroup$ – reuns Aug 11 at 12:10
  • $\begingroup$ Do you follow my answer ? $\endgroup$ – reuns Aug 11 at 13:12
  • For $a > 0$ let $f_a(x) = \log(x-ia)$. With the residue theorem and the fact that $f_a' \in L^2$ we find the Fourier transform of $f_a'(x)=\frac1{x-ia}$ is $\widehat{f_a'} = 2i \pi e^{\omega a}1_{\omega < 0}$.

  • Let $f = \lim_{a \to 0} f_a = \log |x| - i\pi 1_{x < 0}$ then $$i\omega \widehat{f} = \widehat{f'} =\lim_{a \to 0}\widehat{f_a'}= \lim_{a \to 0}2i \pi e^{\omega a}1_{\omega < 0} = 2i \pi 1_{\omega < 0}= i\pi-i \pi \,sign(\omega) $$

  • Let $fp(\frac1{|\omega|})$ and $pv(\frac1\omega)$ be the distributions defined by $$<fp(\frac1{|\omega|}),\phi> = \int_{-\infty}^\infty \frac1{|\omega|} (\phi(\omega)-\phi(0) e^{-\omega^2})d\omega\\ <pv(\frac1{\omega}),\phi> = \int_{-\infty}^\infty \frac1{\omega} (\phi(\omega)-\phi(0) e^{-\omega^2})d\omega$$

  • Then $i\omega (\pi pv(\frac1\omega)-\pi \, fp(\frac1{|\omega|}) =i\pi-i\pi \,sign(\omega)$ so that $\omega( \widehat{f}-(\pi pv(\frac1\omega)-\pi \, fp(\frac1{|\omega|})) = 0 $ which means $$\widehat{f} = \pi pv(\frac1\omega)-\pi \, fp(\frac1{|\omega|}) + c \delta(\omega)$$

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