What is the fourier transform of $e^{2t-t^2}$? I have tried to solve the transform for this exponential but haven't been able to work out how to do it. I know the transform for $e^{-t^2}$ but coupled with the positive term it becomes very difficult to solve.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{-\infty}^{\infty}\expo{2t - t^{2}}\expo{\ic\omega t}\,\dd t & =
\int_{-\infty}^{\infty}\exp\pars{-\bracks{t^{2} - \pars{2 + \omega\ic }t}}
\,\dd t
\\[5mm] & =
\int_{-\infty}^{\infty}\exp\pars{-\braces{t -
\bracks{1 + {1 \over 2}\,\omega\ic}}^{\,2} +
\bracks{1 + {1 \over 2}\,\omega\ic}^{\,2}}\,\dd t
\\[5mm] & =
\exp\pars{\bracks{1 + {1 \over 2}\,\omega\ic }^{\,2}}
\int_{-\infty - \omega\ic/2}^{\infty - \omega\ic/2}\label{1}\tag{1}
\\[5mm] & =
\exp\pars{\bracks{1 + {1 \over 2}\,\omega\ic }^{\,2}}\
\overbrace{\int_{-\infty}^{\infty}
\expo{-t^{2}}\,\dd t}^{\ds{\root{\pi}}}\label{2}\tag{2}
\\[5mm] & =
\bbx{\exp\pars{\bracks{1 + {1 \over 2}\,\omega\ic }^{\,2}}\
\root{\pi}}
\end{align}

From \eqref{1} to \eqref{2}, I performed a contour deformation to restore the integration to the real axis.

