Characteristic function of $X^2$ where $X: \mathcal N(0,1)$. $$\int_{-\infty}^{+ \infty} e^{itx^2}\frac{1}{2 \pi}e^{-\frac{x^2}{2}}dx?$$
I just need to solve this integral. But, I don't know how. When finding the characteristic of $X$ i would write $e^{itX}=\cos tX+ \sin tX$ and then use the fact that $e^{-\frac{x^2}{2}}$ is even and $\sin tX$ is odd, over a symetric domain in the integral, making that $0$. But in this case I do not have that. Help on this integral ?
When $z$ is a positive real number we have
$$f(z) = \int_{-\infty}^\infty e^{-zt^2}{\rm d}t = \frac{1}{\sqrt{z}}\int_{-\infty}^\infty e^{-u^2}{\rm d}u = \sqrt{\frac{\pi}{z}}$$
where we have used a substitution $u=\sqrt{z}t$ and the result for the Gaussian integral whose derivation can be found here.
The integral in the question corresponds to $\frac{1}{2\pi} f\left(\frac{1}{2}-it\right)$ so to solve your problem we need to know $f(z)$ for complex $z$. The short answer is that the formula above holds for complex $z$ as long as Re$(z)>0$ (see proof below), so your integral evaluates to $$\frac{1}{\sqrt{2\pi(1-2it)}} = \frac{\cos (\theta )-2 t \sin (\theta )+i (\sin (\theta )+2 t \cos (\theta ))}{\sqrt{2\pi } \left(4 t^2+1\right)^{\frac{3}{4}}}$$ where $\theta = \frac{\arctan(-2t)}{2}$.
Let $\gamma$ be a rectangle in the right half plane. We have $|e^{-zt^2}| = e^{-t^2\text{Re}(z)}$ so $\int |e^{-zt^2}|{\rm d}t < \infty$ for all $z$ with Re$(z) > 0$ so by Fubini's theorem $$\oint_\gamma f(z){\rm d}z = \oint_\gamma \left[\int_{-\infty}^\infty e^{-zt^2}{\rm d}t\right]{\rm d}z = \int_{-\infty}^\infty\left[\oint_\gamma e^{-zt^2}{\rm d}z\right]{\rm d}t = 0$$ Since $\gamma$ was arbitrary Morera's theorem tell us that $f(z)$ is analytic in the right half plane and since $f(z) = \sqrt{\frac{\pi}{z}}$ for $z\in\mathbb{R}_+$ and both $f(z)$ and $\sqrt{\frac{\pi}{z}}$ is analytic in the right half plane the identity theorem gives us
$$f(z) = \sqrt{\frac{\pi}{z}}~~~\text{for all}~~~z\in\mathbb{C}~~~\text{with}~~~\text{Re}(z) > 0$$
