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 ? 

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.
