Fourier transform of $\operatorname{erfc}^2\left|x\right|$ Could you please help me to find the Fourier transform of
$$f(x)=\operatorname{erfc}^2\left|x\right|,$$
where $\operatorname{erfc}z$ denotes the the complementary error function.

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$$
\mbox{Let's}\quad \phi_{1}\pars{k} \equiv \int_{-\infty}^{\infty}{\rm erfc}\pars{\verts{x}}\expo{-\ic kx}\,\dd x
$$
such that
\begin{align}
\phi_{2}\pars{k}&=
\int_{-\infty}^{\infty}{\rm erfc}^{2}\pars{\verts{x}}\expo{-\ic kx}\,\dd x
=\int_{-\infty}^{\infty}
\int_{-\infty}^{\infty}\phi_{1}\pars{p}\expo{\ic px}\,{\dd p \over 2\pi}
\int_{-\infty}^{\infty}\phi_{1}^{*}\pars{q}\expo{-\ic qx}\,{\dd q \over 2\pi}\,
\expo{-\ic kx}\,\dd x
\\[3mm]&=
{1 \over 2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\dd p\,\dd q\,
\phi_{1}\pars{p}\phi_{1}^{*}\pars{q}\
\overbrace{\int_{-\infty}^{\infty}\expo{\ic\pars{p - q - k}x}\,{\dd x \over 2\pi}}
^{\ds{\delta\pars{p - q - k}}}
\end{align}

$$
\phi_{2}\pars{k} = {1 \over 2\pi}\int_{-\infty}^{\infty}
\phi_{1}\pars{p}\phi_{1}^{*}\pars{p - k}\,\dd p
$$

\begin{align}
\phi_{1}\pars{k}&=\int_{-\infty}^{\infty}{2 \over \root{\pi}}
\int_{\verts{x}}^{\infty}\dd t\,\expo{-t^{2}}\expo{-\ic kx}\,\dd x
={2 \over \root{\pi}}
\int_{-\infty}^{\infty}\dd t\,\expo{-t^{2}}
\int_{-\infty}^{\infty}\Theta\pars{t - \verts{x}}\expo{-\ic kx}\,\dd x
\\[3mm]&={2 \over \root{\pi}}
\int_{-\infty}^{\infty}\dd t\,\expo{-t^{2}}\Theta\pars{t}
\int_{-t}^{t}\expo{-\ic kx}\,\dd x
={2 \over \root{\pi}}
\int_{0}^{\infty}\dd t\,\expo{-t^{2}}{\sin\pars{kt} \over k}
={2 \over \root{\pi}}\,{{\rm F}\pars{k/2} \over k}
\end{align}
where $\ds{{\rm F}\pars{z}}$ is the Dawson Integral.

$$\color{#00f}{\large%
\phi_{2}\pars{k} = {2 \over \pi^{2}}
\int_{-\infty}^{\infty}{{\rm F}\pars{p/2}{\rm F}\pars{\bracks{p - k}/2}
\over k\pars{p - k}}\,\dd p}
$$

