Show that the normal distribution reduces to \(f(x)=\frac{e^{-(x / a)^{2}}}{a \sqrt{\pi}}\) if it is centered at the origin, where \(a=\sqrt{2} \sigma\). The maximum of this function occurs at \(f(0)=\frac{1}{a \sqrt{\pi}}\), where \(a\) is the width of the curve at half the maximum value.

(a) Show that the Fourier transform of this function is \(F(w)=\mathscr{F}\{f(x)\}=\) \(\frac{1}{\sqrt{2 \pi}} e^{(-w / b)^{2}}\). This function is also a Gaussian function, where \(b=2 / a\) is the corresponding width of the transformed curve at half-maximum.

(b) Recover \(f(x)\) by evaluating the inverse Fourier transform \(\mathscr{F}^{-1}\{F(w)\}\) for the function \(F(w)\) given in part (a).