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

Question

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.

Answer 1

• 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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Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{I}\pars{\omega} & = \lim_{\epsilon \to 0^{+}}\int_{-\infty}^{\infty} \ln\pars{\ic\bracks{x - \ic\epsilon}}\expo{\ic\omega x}\,\dd x \\[5mm] & \stackrel{x\ =\ -\ic s}{=}\,\,\, \lim_{\epsilon \to 0^{+}}\int_{-\infty\ic}^{\infty\ic} \ln\pars{s + \epsilon}\expo{\omega s}\pars{-\ic}\,\dd s \\[5mm] & = -\ic\lim_{\epsilon \to 0^{+}}\bracks{\expo{-\omega\epsilon} \int_{\epsilon - \infty\ic}^{\epsilon + \infty\ic} \ln\pars{s}\expo{\omega s}\,\dd s} = -\ic\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} \ln\pars{s}\expo{\omega s}\,\dd s \\[5mm] & = -\ic\bracks{\omega > 0} \\[2mm] & \times \bracks{% -\int_{-\infty}^{0}\bracks{\ln\pars{-s} + \ic\pi}\expo{\omega s}\dd s - \int_{0}^{-\infty}\bracks{\ln\pars{-s} - \ic\pi}\expo{\omega s}\dd s} \\[5mm] & = -\ic\bracks{\omega > 0} \\[2mm] & \times\bracks{% -\int_{0}^{\infty}\bracks{\ln\pars{s} + \ic\pi}\expo{-\omega s}\dd s + \int_{0}^{\infty}\bracks{\ln\pars{s} - \ic\pi}\expo{-\omega s}\dd s} \\[5mm] & = -2\pi\bracks{\omega > 0}\int_{0}^{\infty}\expo{-\omega s}\dd s = \bbx{-2\pi\,{\bracks{\omega > 0} \over \omega}} \\ & \end{align}