Evaluating an integral of a form related to $\int_{-\infty}^{\infty} e^{-ax^2} \cdot e^{-2\pi i k x} dx$ Claim
\begin{equation} 
\int_{\mathbb R} \exp\left(-2\pi \cdot \left(\frac{x}{\sqrt 2}\right)^2 \right) \cdot \exp\left(-2i \pi \frac{x}{\sqrt 2} \cdot f\right) \mathrm{d}\left( \frac{x}{\sqrt{2}} \right)
= \frac{1}{\sqrt{2}} \exp(-\pi f^{2}/2).
\end{equation}
where I cannot get anyway the second power in $f$. 
Wolfram alpha also returns correct result here.
However, I do not understand how. 
The general formula of Fourier transform of Gaussian is
\begin{equation}
\int_{-\infty}^{\infty} e^{-ax^2} \cdot e^{-2\pi i k x} dx = F_{x}[ e^{-ax^2} ]
\end{equation}
Looking at (1), I cannot see how the $f$ squares in the parameter of the result because $x := (x/\sqrt{2})$ in (1), and $f$ term apparently corresponds $2\pi i k$. 
I think the general equation from Wolfram does not apply here directly. 
Using $a \mapsto 2\pi$, $x \mapsto x/\sqrt{2}$, and $k \mapsto f$. 
I really get the General formula there. 
Solving the Fourier transform
\begin{equation}
F_{x} [e^{-ax^2}] = F_{x} [e^{-2\pi x^2}]
\end{equation}
where missing the $f$ term. 
So I am still thinking how the $f$ term comes there in the end. 
If $x$ was also mapped to $f$, there would be $f$ term but then the original equation cannot hold.

How can you get the result of integration?

Hint:
Let$$I=\int_{\mathbb R}e^{-x^2/2}\cos(fx)dx.$$
Then deriving on $f$ and integrating by parts,
$$I'_f=-\int_{\mathbb R}xe^{-x^2/2}\sin(fx)dx=\left.e^{-x^2/2}\sin(fx)\right|_{\mathbb R}-\int_{\mathbb R} fe^{-x^2/2}\cos(fx)dx=-fI.$$
The solution of this differential equation is
$$I=Ce^{-f^2/2}.$$
