Fourier transform of Cauchy kernel

Question

For a fixed $z\in \mathbb{C}\setminus\mathbb{R}$, define the function \begin{align*} C:\mathbb{R}&\to\mathbb{C} \\ x&\mapsto \frac{1}{2\pi i(x-z)} \end{align*}

I have already shown that $C(x)\in L^2(\mathbb{R})$, and now I am trying to compute the Fourier transform of $C$

$$\mathcal{F}[C(x)](p):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-ipx}C(x)dx$$ and it seems hard.. I think it will probably need to be in the sense of a principle value, but I'm a bit stuck on how to proceed.

Answer 1

The main idea of the following is the fact that the antiderivative of $t\mapsto e^{-izt}$ is equal ${-1\over iz}e^{-izt}$ for nonreal $z.$

For ${\rm Im}\,z>0$ consider the function $$f_z(p)=\begin{cases} e^{-izp} & p\le 0\\ 0 & p>0\end{cases}$$ The function $f_z$ is in $L^1\cap L^2.$ The inverse Fourier transform of $f_z$ is equal $$\mathcal{F}^{-1}(f_z)(x)={1\over \sqrt{2\pi}}\int\limits_{-\infty}^0e^{-izp}e^{ipx}\,dp={1\over \sqrt{2\pi}}{1\over i(x-z) }$$ Similarly for ${\rm Im}\,z<0$ and $$f_z(p)=\begin{cases} e^{-izp} & p\ge 0\\ 0 & p<0\end{cases}$$ we get $$\mathcal{F}^{-1}(f_z)(x)=-{1\over \sqrt{2\pi}}{1\over i(x-z) } $$ As $(\mathcal{F}\circ \mathcal{F}^{-1})(f_z)=f_z,$ we get $\mathcal{F}(g_z)=f_z,$ where $g_z(x) =\sqrt{2\pi}\,(2\pi i)^{-1}(x-z)^{-1}.$

Answer 2

Let $C(x)=\frac1{x-z}$, $z\in\mathbb{C}\setminus\mathbb{R}$. Then, we have

$$\begin{align} \mathscr{F}\{C\}(p)&=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty C(x)e^{-ipx}\,dx\\\\ &=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \frac1{x-z}\,e^{-ipx}\,dx \end{align}$$

We can evaluate this integral using the resiude theorem.

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For $\text{Im}(p>0)$, we close the contour in the lower-half plane. If $\text{Im}(z)>0$, then the integral is zero. If $\text{Im}(z)<0$, then we see that

$$\begin{align} \mathscr{F}\{C\}(p)&=-2\pi i \frac1{\sqrt{2\pi}}\text{Res}\left(\frac{e^{-ixp}}{x-z},x=z\right)\end{align}=-i\sqrt{2\pi}e^{-izp}$$

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For $\text{Im}(p<0)$, we close the contour in the upper-half plane. If $\text{Im}(z)<0$, then the integral is zero. If $\text{Im}(z)>0$, then we see that

$$\begin{align} \mathscr{F}\{C\}(p)&=2\pi i \frac1{\sqrt{2\pi}}\text{Res}\left(\frac{e^{-ixp}}{x-z},x=z\right)\end{align}=i\sqrt{2\pi}e^{-izp}$$

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