Gaussians as good kernels

Question

Theorem. If $\delta>0$ and $$K_{\delta}(x) = \delta^{-\frac{1}{2}}e^{- \pi \frac{x^2}{\delta}},$$ then $$\widehat{K_{\delta}}(\xi) = e^{- \pi \delta \xi^2}.$$ Proof.
I started the proof in a very expected way. $$\widehat{K_{\delta}}(\xi) = \int \limits_{-\infty}^{\infty} \delta^{-\frac{1}{2}}e^{- \pi \frac{x^2}{\delta}} e^{-2 \pi ix \xi} \mbox{d}x. \tag{1}$$ Unfortunately I don't know how to evaluate $(1)$. I think using the following equality might help $$\int \limits_{-\infty}^{\infty} e^{- \pi x} \mbox{d}x =1$$ but I don't know how to use it. I would appreciate any hints or tips.

Answer 1

The integral is equal to $$ \int _{-\infty}^{\infty} \delta^{-\frac{1}{2}}e^{- \pi \frac{x^2}{\delta}} e^{-2 \pi ix \xi}dx = \int _{-\infty}^{\infty}e^{- \pi x^2} e^{-2 \pi i\sqrt{\delta} x \xi}dx= e^{-\pi\delta\xi^2}\int _{-\infty}^{\infty}e^{- \pi (x+i\sqrt{\delta}\xi)^2} dx. $$ Note that $$ \int _{-\infty}^{\infty}e^{- \pi (x+i\sqrt{\delta}\xi)^2} dx=\int_{\Im(z)=\sqrt{\delta}\xi}e^{- \pi z^2} dz. $$ By Cauchy's integral theorem, we can show that $$ \int_{\Im(z)=\sqrt{\delta}\xi}e^{- \pi z^2} dz =\int_{\Im(z)=0}e^{- \pi z^2} dz=\int_{-\infty}^\infty e^{- \pi z^2} dz=1. $$