Theorem: Fourier Transform of Normal Function

Let $\map f x$ be defined as $\sqrt \pi$ times the normal probability density function where $\mu = 0$ and $\sigma = \dfrac {\sqrt 2} 2$:
{{begin-eqn}}
{{eqn | l = \map f x
      | r = \dfrac {\sqrt {\pi} } {\dfrac {\sqrt 2} 2 \sqrt {2 \pi} } \map \exp {-\dfrac {\paren {x - 0}^2} {2 \paren {\dfrac {\sqrt 2} 2 }^2} }
      | c = 
}}
{{eqn | r = e^{-x^2}
      | c = 
}}
{{end-eqn}}
Then:
:$\map {\hat f} s = \sqrt \pi e^{-\paren {\pi s }^2}$
where $\map {\hat f} s$ is the Fourier transform of $\map f x$.

Proof:
By the definition of a Fourier transform:
{{begin-eqn}}
{{eqn | l = \map {\hat f} s
      | r = \int_{-\infty}^\infty e^{-2 \pi i x s} \map f x \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty e^{-2 \pi i x s} e^{-x^2} \rd x
      | c = 
}}
{{end-eqn}}
Taking the derivative with respect to $s$, we have:
{{begin-eqn}}
{{eqn | l = \dfrac \d {\d s} \map {\hat f} s
      | r = \dfrac \d {\d s} \int_{-\infty}^\infty e^{-2 \pi i x s} e^{-x^2} \rd x
      | c = 
}}
{{eqn | r = -2 \pi i \int_{-\infty}^\infty e^{-2 \pi i x s} x e^{-x^2} \rd x
      | c = 
}}
{{end-eqn}}
Integrating by parts, we have:
Let $u = e^{-2 \pi i x s}$ and $\rd v = x e^{-x^2} \rd x$
Then:
:$\rd u = -2 \pi i s e^{-2 \pi i x s} \rd x$ and $v = -\dfrac 1 2 e^{-x^2}$
Hence:
{{begin-eqn}}
{{eqn | l = \dfrac \d {\d s} \map {\hat f} s
      | r = -2 \pi i \paren {\paren {\lim_{\gamma \mathop \to \infty} \intlimits {-\dfrac 1 2 e^{-2 \pi i x s} e^{-x^2} } {-\gamma} {+\gamma} } - \int_{-\infty}^\infty \paren {-\dfrac 1 2 e^{-x^2} } \paren {-2 \pi i s e^{-2 \pi i x s} } \rd x}
      | c = 
}}
{{eqn | r = -2 \pi i \paren {0 - \int_{-\infty}^\infty \paren {-\dfrac 1 2 e^{-x^2} } \paren {-2 \pi i s e^{-2 \pi i x s} } \rd x}
      | c = first part $\paren {u v}$ sums to zero
}}
{{eqn | r = -2 \pi i \paren {-\pi i \int_{-\infty}^\infty \paren {e^{-x^2} } \paren { s e^{-2 \pi i x s} } \rd x}
      | c = 
}}
{{eqn | r = 2 \pi^2 i^2 \int_{-\infty}^\infty \paren {e^{-x^2} } \paren {s e^{-2 \pi i x s} } \rd x
      | c = 
}}
{{eqn | r = -2 \pi^2 \int_{-\infty}^\infty \paren {e^{-x^2} } \paren {s e^{-2 \pi i x s} } \rd x
      | c = $i^2 = -1$
}}
{{end-eqn}}
We now have the following:
{{begin-eqn}}
{{eqn | l = \frac \d {\d s} \map {\hat f} s 
      | r = \int_{-\infty}^\infty \paren {-2 \pi^2 s e^{-x^2} } e^{-2 \pi i x s} \rd x
      | c = 
}}
{{eqn | r = -2 \pi^2 s \map {\hat f} s
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \d {\map {\hat f} s}
      | r = -2 \pi^2 s \map {\hat f} s \rd s
      | c = 
}}
{{eqn | l = \dfrac {\map \d {\map {\hat f} s } } {\map {\hat f} s}
      | r = -2 \pi^2 s \rd s
      | c = Solution to Separable Differential Equation
}}
{{eqn | l = \int \dfrac {\map \d {\map {\hat f} s } } {\map {\hat f} s}
      | r = \int -2 \pi^2 s \rd s
      | c = integrating
}}
{{eqn | l = \map \ln {\map {\hat f} s}
      | r = -\pi^2 s^2 + C
      | c = Primitive of Function under its Derivative, Primitive of Power
}}
{{eqn | l = \map {\hat f} s 
      | r = A e^{-\pi^2 s^2}
      | c = Exponential of Natural Logarithm: $A = e^C$
}}
{{end-eqn}}
We solve for $A$ by setting $s = 0$:
{{begin-eqn}}
{{eqn | l = \map {\hat f} 0
      | r = \int_{-\infty}^\infty e^{-2 \pi i x 0} e^{-x^2} \rd x
      | c = 
}}
{{eqn | r = \int_{-\infty}^\infty e^{-x^2} \rd x
      | c = Exponential of Zero
}}
{{eqn | r = \sqrt \pi
      | c = Gaussian Integral
}}
{{end-eqn}}
Therefore:
:$\map {\hat f} s = \sqrt \pi e^{-\paren {\pi s}^2}$
{{qed}}