Fourier transform of $G_s(x) = \frac{1}{\sqrt{s}} e^{-\frac{\pi x^2}{s}}$ I have to calculate the Fourier transform of $G_s(x) = \dfrac{1}{\sqrt{s}} e^{-\dfrac{\pi x^2}{s}}$; $s > 0$.
I have proven that $G_s$ is a mollifier when $s \rightarrow 0^+$.
Have you an idea to calculate this transform ? 
 A: HINT:
Enforce the substitution $s\mapsto s^2$ to find
$$I=\int_0^\infty \frac{e^{-\pi x^2/s}}{\sqrt s}e^{iks}\,ds=2\int_0^\infty e^{-(\pi x^2/s^2 -iks^2)}\,ds$$
Find constants $a$ and $b$ such that
$$I =\int_{-\infty}^\infty e^{-a\left(\left(\frac{b}{s}\right)^2+\left( \frac{s}{b}\right)^2 \right)}\,ds$$
Substitute $s/b\mapsto s$ and deform the transformed integration path back onto the real line.
Note that $\left(\frac 1s\right)^2+s^2=\left(s-\frac1s\right)^2+2$.
Finally, exploit the Cauchy-Schlomilch transform, which is a special case of Glasser's Master Theorem.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\int_{0}^{\infty}{\expo{-\pi x^{2}/s} \over \root{s}}\,\expo{\ic ks}\,\dd s & =
\int_{0}^{\infty}\exp\pars{-\,{\pi x^{2} \over s} + \ic ks}
{\dd s \over \root{s}}
\end{align}

Set $\ds{s = A\expo{\theta}}$. I'll choose $\ds{A}$ in a 'convenient way' later on. Namely,

\begin{align}
\int_{0}^{\infty}{\expo{-\pi x^{2}/s} \over \root{s}}\,\expo{\ic ks}\,\dd s & =
\int_{0}^{\infty}\exp\pars{-\pi x^{2}A^{-1}\expo{-\theta} - \bracks{-\ic k} A\expo{\theta}}
{A\expo{\theta}\dd\theta \over \root{A\expo{\theta}}}
\end{align}

I'll choose $\ds{A}$ with $\ds{\pi x^{2}A^{-1} = -\ic kA \implies A =
\root{\pi x^{2} \over -\ic k} = \root{\pi \over k}\verts{x}\expo{\ic\pi/4}}$. Then,

\begin{align}
&\int_{0}^{\infty}{\expo{-\pi x^{2}/s} \over \root{s}}\,\expo{\ic ks}\,\dd s
\\[5mm]  = &\
\int_{-\infty A}^{\infty A}
\exp\pars{-\ic k
\bracks{\root{\pi \over k}\verts{x}\expo{\ic\pi/4}}\bracks{2\cosh\pars{\theta}}}
\pars{\pi \over k}^{1/4}\root{\verts{x}}\expo{\ic\pi/8}\expo{\theta/2}
\,\dd\theta
\\[5mm]  = &\
\pars{\pi \over k}^{1/4}\root{\verts{x}}\expo{\ic\pi/8}
\int_{-\infty A}^{\infty A}\exp\pars{-2\ic
\root{\pi k}\verts{x}\expo{\ic\pi/4}\cosh\pars{\theta}}\expo{\theta/2}
\,\dd\theta
\\[5mm]  = &\
\pars{\pi \over k}^{1/4}\root{\verts{x}}\expo{\ic\pi/8}
\int_{0}^{\infty A}\exp\pars{-2\ic\root{\pi k}\verts{x}\expo{\ic\pi/4}
\bracks{2\sinh^{2}\pars{\theta \over 2} + 1}}
\cosh\pars{\theta \over 2}\,\dd\theta
\end{align}

With $\ds{t \equiv \sinh\pars{\theta/2}}$:

\begin{align}
&\int_{0}^{\infty}{\expo{-\pi x^{2}/s} \over \root{s}}\,\expo{\ic ks}\,\dd s
\\[5mm]  = &\
\pars{\pi \over k}^{1/4}\root{\verts{x}}\expo{\ic\pi/8}
\bracks{\exp\pars{-2\ic\root{\pi k}\verts{x}\expo{\ic\pi/4}}}\ \times
\\[2mm] &\
\lim_{R \to \infty}\int_{0}^{\sinh\pars{RA/2}}
\exp\pars{-4\ic\root{\pi k}\verts{x}\expo{\ic\pi/4}t^{2}}\,2\,\dd t
\end{align}

I guess you can continue from the last expression.

A: \begin{align}
    \widehat{G_s}(y)&= \frac{1}{\sqrt{2\pi s}} \int_{-\infty}^{\infty}e^{-\pi x^2/s}e^{-iyx}dx \\
    \widehat{G_s}'(y)&=\frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{\infty}-ixe^{-\pi x^2/s}e^{-iyx}dx \\
   & = \frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{\infty}\frac{is}{2\pi}\frac{d}{dx}(e^{-\pi x^2/s})e^{-iyx}dx \\
   & = -\frac{1}{\sqrt{2\pi s}}\frac{is}{2\pi}\int_{-\infty}^{\infty}e^{-\pi x^2/2}\frac{d}{dx}e^{-iyx}dx \\
   & = -\frac{1}{\sqrt{2\pi s}}\frac{sy}{2\pi}\int_{-\infty}^{\infty}e^{-\pi x^2/2}e^{-iyx}dx \\
   & = -\frac{sy}{2\pi}\widehat{G_s}(y).
\end{align}
This is a first order ODE in $s$ with solution
$$
        \widehat{G_s}(y)=A(s)e^{-sy^2/4\pi}
$$
$A(s)$ is determined by setting $y=0$:
$$
     \widehat{G_s}(0)=A(s) \\
      A(s) = \frac{1}{\sqrt{2\pi s}}\int_{-\infty}^{\infty}e^{-\pi x^2/s}dx
$$
You can compute $A(s)$; I believe it is constant. This is a standard normal distribution (Guassian) integral.
