Just for my own curiosity, I'm trying to derive the characteristic function of $X\sim\chi^2_1$, the $\chi^2$ distribution with 1 degree of freedom. According to wikipedia, it is $$ E[e^{itX}]=\frac{1}{\sqrt{1-2it}},\quad t\in\mathbb{R}. $$ To proceed, I can think of $X$ as $U^2$ where $U\sim N(0,1)$. Then $$ E[e^{itX}]=E[e^{itU^2}]=\int_{-\infty}^{+\infty}e^{itu^2}\frac{1}{2\pi}e^{-u^2/2}du=2\int_0^{+\infty}e^{itu^2}\frac{1}{2\pi}e^{-u^2/2}du. $$ Let's make the substitution $y=u^2$ so that $dy=2udu$ with $u=\sqrt{y}$ for $u\geq 0$. Then $$ E[e^{itX}]=2\int_0^\infty e^{ity}e^{-y/2}\frac{dy}{2\sqrt{2\pi y}}=\int_0^\infty\cos(ty)e^{-y/2}\frac{dy}{\sqrt{2\pi y}}+i\int_0^\infty\sin(ty)e^{-y/2}\frac{dy}{\sqrt{2\pi y}} $$ but I can't proceed any further. :(
I have seen a derivation here but there is a substitution step that involves $i$ so I'm not comfortable with using it.