Inverse Fourier transform of a rational function

Question

I was wondering how I can solve the inverse Fourier transform of $x/(1+x^{2})$ There isn't anything quite similar to this in the table of Fourier transforms that I have seen and I am not quite sure how to do it.

Any help is appreciated.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \left\{\begin{array}{rcl} \ds{\mrm{f}\pars{x}} & \ds{\equiv} & \ds{{x \over x^{2} + 1} = \int_{-\infty}^{\infty} \hat{\mrm{f}}\pars{k}\expo{\ic kx}\,{\dd k \over 2\pi}} \\[2mm] \ds{\hat{\mrm{f}}\pars{k}} & \ds{=} & \ds{\int_{-\infty}^{\infty}{x \over x^{2} + 1} \expo{-\ic kx}\,\dd x} \end{array}\right. $$

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\begin{align} \hat{\mrm{f}}\pars{k} & = \int_{-\infty}^{\infty}{x \over x^{2} + 1} \,\expo{-\ic kx}\dd x = -\ic\int_{-\infty}^{\infty}{x\sin\pars{kx} \over x^{2} + 1}\,\dd x \\[5mm] & = -\ic\,\mrm{sgn}\pars{k} \int_{-\infty}^{\infty}{x\sin\pars{\verts{k}x} \over x^{2} + 1}\,\dd x \\[5mm] & = -\ic\,\mrm{sgn}\pars{k}\,\Im \int_{-\infty}^{\infty}{x\expo{\ic\verts{k}x} \over x^{2} + 1}\,\dd x \\[5mm] & = -\ic\,\mrm{sgn}\pars{k}\,\Im \pars{2\pi\ic\,{\ic\expo{\ic\verts{k}\ic} \over 2\ic}} \\[5mm] & = \bbx{-\pi\,\mrm{sgn}\pars{k}\expo{-\verts{k}}\ic} \\ & \end{align}