Inverse fouriertransform I'm trying to take an inverse fourier transform of an exponential function (the $i$ is the complex number):
$\mathcal{F}^{-1} \big\lbrace\exp \big(\frac{-\hbar}{2m}(2\pi k)^2 it\big)\big\rbrace = \int_{-\infty}^{\infty} \exp \big(\frac{-\hbar}{2m}(2\pi k)^2 it\big) \exp(2\pi ikx) dk$
In my solution I have a $t$ in the denominator under a square root:
$\exp \big(\frac{mx^2i}{2\hbar t}\big) \sqrt{\frac{m}{2\pi it\hbar}}$
I tried to solve this by getting a product of the type $(k+b)^2$ in the exponent:
$ \int_{-\infty}^{\infty} \exp \big\lbrace\frac{-2\pi^2 it\hbar}{m}\big(k - \frac{mx}{2\hbar \pi t}\big)^2\big\rbrace \exp \big(\frac{mx^2i}{2\hbar t}\big) dk$
A $t$ in the denominator forms a problem, because at time $= 0$ I should get an initial condition (so my solution should by one so a prefactor say $A$ is the solution for time $= 0$). Can someone help me? Thank you.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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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\begin{align}
&\bbox[10px,#ffd]{\ds{%
\int_{-\infty}^{\infty}
\exp\pars{-\,{\hbar \over 2m}\bracks{2\pi k}^{\,2}\,\ic t}
\exp\pars{2\pi\ic kx}\,\dd k}} =
\int_{-\infty}^{\infty}\exp\pars{-\,{2\pi^{2}\hbar\ic t \over m}
\bracks{k^{2} - {mx \over \pi\hbar t}\,k}}\,\dd k
\\[5mm] = &\
\int_{-\infty}^{\infty}\exp\pars{-\,{2\pi^{2}\hbar\ic t \over m}
\braces{\bracks{k - {mx \over 2\pi\hbar t}}^{2} - {m^{2}x^{2} \over 4\pi^{2}\hbar^{2}t^{2}}}}\,\dd k
\\[5mm] = &\
\exp\pars{{mx^{2} \over 2\hbar t}\,\ic}\
\underbrace{\qquad\int_{-\infty}^{\infty}
\exp\pars{-\,{2\pi^{2}\hbar\ic t \over m}k^{2}}\,\dd k\qquad}
_{\ds{Fresnel\ Integral\ =\ \root{\pi \over \ic\pars{2\pi^{2}\hbar\ic t}/m}}}\ =\
\bbx{\exp\pars{{mx^{2} \over 2\hbar t}\,\ic}\root{-\,{m \over 2\pi\hbar t}}}
\end{align}
