What is the fourier transform of $e^{2t-t^2}$?

Question

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.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.