# Fourier transform of $\log(i (x-i\epsilon))$

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.

• $- 2\pi \log(\omega) \delta(\omega) - 2\pi \Theta(\omega) \frac{1}{\omega}$ doesn't make any sense, you should not follow a paper mentioning it. – reuns Aug 11 at 11:36
• @reuns Ok I deleted the expression. It is also not exactly whats written in the paper. Does the first equality make sense? – lomby Aug 11 at 11:54
• No, your integral doesn't converge so it doesn't make sense to use contour integration. – reuns Aug 11 at 12:10
• Do you follow my answer ? – 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 $$ = \int_{-\infty}^\infty \frac1{|\omega|} (\phi(\omega)-\phi(0) e^{-\omega^2})d\omega\\ = \int_{-\infty}^\infty \frac1{\omega} (\phi(\omega)-\phi(0) e^{-\omega^2})d\omega$$
• 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 $$\widehat{f} = \pi pv(\frac1\omega)-\pi \, fp(\frac1{|\omega|}) + c \delta(\omega)$$