Joint characteristic function of $x$ and $y=x^2$ if $x$ is the standard normal variable

Question

How to find the joint characteristic function of $x$ and $y=x^2$ if $x$ is standard normal variable with mean $0$ and variance $1$?

Answer 1

$$ f(t,s)=\langle e^{\mathrm{i}(tx+sx^2)}\rangle=\int_{-\infty}^\infty \frac{dx}{\sqrt{2\pi}}e^{-x^2/2}e^{\mathrm{i} (tx+s x^2)}=\int_{-\infty}^\infty \frac{dx}{\sqrt{2\pi}}e^{-x^2(1/2-\mathrm{i}s)}e^{\mathrm{i}tx}\ , $$ and setting $x\sqrt{1/2-\mathrm{i}s}=z$ $$ f(t,s)=\frac{1}{\sqrt{1/2-\mathrm{i}s}}\int_{-\infty}^\infty \frac{dz}{\sqrt{2\pi}}e^{-z^2}e^{\mathrm{i}\frac{t}{\sqrt{1/2-\mathrm{i}s}}z}=\frac{e^{-t^2/(2-4\mathrm{i}s)}}{\sqrt{1-2\mathrm{i}s}}\ . $$