Fourier transform of $e^{-\frac{1}{2}x^2}$ [duplicate]

Fourier transform of $$e^{-\frac{1}{2}x^2}$$

I know to use the Fourier transform direct formula however I keep getting an algebraic mess. so if someone could help me out with detailed steps that would be very much appreciated!

• Let $F(s) = \int_{-\infty}^\infty e^{-x^2/2 - sx}dx$. For $s \in \mathbb{R}$ you can apply the change of variable $y = x+s$. What do you get ? – reuns Aug 16 '17 at 9:53

1 Answer

\begin{align} \hat{f}(k)=&\int_\mathbb R \exp\left(- \frac 1 2 x^2 - i k x\right) \, dx \\ =& \int \exp\left(- \frac 1 2 (x^2 + 2i k x)\right) \, dx \\ =&\int \exp\left(- \frac 1 2 (x + i k )^2-\frac 1 2 k^2\right) \, dx \\ =&e^{- \frac 1 2 k^2}\int \exp\left(- \frac 1 2 y^2\right) \, dy\\ =&\sqrt{2 \pi}e^{- \frac 1 2 k^2} \end{align}

You possibly defined the Fourier transform as a the product of this with $2 \pi$ or $\sqrt{2 \pi}$.
You can think of the fourth equality sign as substituting $y=x+ik$. This is justified by using contour integration in the complex plane using a rectangular path with one path along the reals, $2$ vertical paths at infinity parallel to the imaginary axis and a path in negative direction along $\mathbb R +ik$.