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

\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}
