S1.

Solution:
As we know, the Fourier transformation may be written by the following way:

$ f(x) = \mathfrak{F}^{-1}\left\{F(j\omega)\right\} = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(j\omega)e^{j\omega x}d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-\omega^{2}T^{2}}e^{j\omega x} d\omega

Let’s apply the Euler’s formula now!

$ f(x)=2\frac{1}{2\pi}\int_{0}^{\infty}e^{-\omega^{2}T^{2}}\text{cos}(\omega x)d\omega = \frac{1}{\pi} \int_{0}^{\infty}e^{-T^{2}\omega^{2}}\text{cos}(\omega x)d\omega; where we have to use the available condition with the following substitutions:
$x_{0}=\omega ; T^{2}=a ; x = b.

Consequently the final solution is:
$f(x)=\frac{1}{\pi} \frac{1}{2} \sqrt{\frac{\pi}{T^{2}}}e^{-(\frac{x}{4T^{2}})^{2}} = \frac{1}{2\sqrt{\pi}T}e^{-\frac{x^{2}}{16T^{4}}

 
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