Show that $\frac{1}{\sqrt{2\pi}} \cdot \left(\int_{- \infty}^{\infty} e^{itx} \cdot e^{-\frac{1}{2}t^2 } dx \right) = e^{-\frac{1}{2}t^2}$ I am struggling to show that $$\frac{1}{\sqrt{2\pi}} \cdot \left(\int_{- \infty}^{\infty} e^{itx} \cdot e^{-\frac{1}{2}x^2 } dx \right) = e^{-\frac{1}{2}t^2}$$
In the same paper we defined  $$F(t):=\left(\int_0^t e^{-x^2}\right)^2 \quad \text{ and} \quad  G(t):=\left( \int _0^1\frac{e^{-t^2(1+x^2)}}{1+x^2} \right)$$
I showed that both functions are differentiable with
 $$F'= 2e^{-t^2} \cdot \int_0^t e^{-x^2}$$ $$G' = -2te^{-t^2} \cdot \int_0^1 e^{-x^2t^2}$$
It follows that $F' + G' = 0$ and $F + G = \frac{\pi}{4}$
With that I showed that $\int_{-\infty}^{\infty} e^{-x^2} = \sqrt{\pi}$.
Any help is appreciated.
 A: By completing the square, we find that 
$$\begin{align}
\int_{-\infty}^\infty e^{-\frac12t^2+itx}\,dt&=e^{-\frac12 x^2}\int_{-\infty}^\infty e^{-\frac12\left(t-ix\right)^2}\,dt\\\\
&=e^{-\frac12 x^2}\int_{-\infty-ix}^{\infty-ix} e^{-\frac12 t^2}\,dt\\\\
\end{align}$$
Then, invoking Cauchy's Integral Theorem, we can deform the contour back onto the real line, which reveals
$$\int_{-\infty-ix}^{\infty-ix} e^{-\frac12 t^2}\,dt=\int_{-\infty}^{\infty} e^{-\frac12 t^2}\,dt=\sqrt {2\pi}$$
Therefore, we have
$$\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12t^2+itx}\,dt=e^{-\frac12 x^2}$$
as was to be shown!
A: Here is another way forward that relies on Feyman's trick.
Let $f$ be defined by the integral of interest
$$f(x)=\int_{-\infty}^\infty e^{-\frac12 t^2}e^{itx}\,dt$$
Differentiating under the integral yields
$$f'(x) =\int_{-\infty}^\infty ite^{-\frac12 t^2}e^{itx}\,dt$$
Integrating by parts with $u=ie^{itx}$ and $v=-e^{-\frac12 t^2}$ reveals
$$f'(x)=-xf(x)$$
from which it is easy to see that $f(x)=Ae^{-\frac12 x^2}$.  Since $f(0)=\sqrt{2\pi}$, we find that $A=\sqrt{2\pi}$ and 
$$\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12 t^2}e^{itx}\,dt=e^{-\frac12x^2}$$
as expected!
