<s>[INST] How do I determine the charasteristic function of $X^2$? I'm wondering how I kind show that the charasteristic function of $X^2$ given that $X\in N(0,1)$ is $\varphi_{X^2}(t)=\frac{1}{\sqrt{1-2it}}$.
I have tried using the change of variables such that $Y=X^2\implies X=\sqrt{Y}$ and then finding the derivative which is $\frac{1}{2\sqrt{y}}$ for using this in the transform formula. Then I apply $y$ instead of $x$ in the standard normal distribution $\phi(x)=\frac{1}{2\pi}e^{\frac{-x^2}{2}}$ and then using that for the charasteristic function formula $\varphi_{X^2}(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ity}e^{\frac{-y}{2}}\frac{1}{2\sqrt{y}}dy$. But this integral gets way too complicated and I think that my method is wrong. Is there anyway I can show that $\varphi_{X^2}(t)=\frac{1}{\sqrt{1-2it}}$?
thanks
 [/INST] $$Y = {X^2} \Rightarrow f\left( y \right) = \frac{1}{{\sqrt {2\pi } }}{y^{ - \frac{1}{2}}}{e^{ - \frac{y}{2}}},x > 0.$$ (because $Y \sim \chi _1^2$) so
    $${\varphi _Y}\left( t \right) = \mathop \int \limits_0^\infty  {e^{ity}} \cdot \frac{1}{{\sqrt {2\pi } }}{y^{ - \frac{1}{2}}}{e^{ - \frac{y}{2}}}dy = \frac{1}{{\sqrt {2\pi } }}\mathop \int \limits_0^\infty  {y^{\frac{1}{2} - 1}}{e^{ - y\left( {\frac{1}{2} - it} \right)}}dy$$
using
$$\mathop \int \limits_0^\infty  {x^{\alpha  - 1}}{e^{ - \beta x}}dx = \frac{{\Gamma \left( \alpha  \right)}}{{{\beta ^\alpha }}}$$, we get
$$ = \frac{1}{{\sqrt {2\pi } }} \cdot \frac{{\Gamma \left( {\frac{1}{2}} \right)}}{{{{\left( {\frac{1}{2} - it} \right)}^{\frac{1}{2}}}}} = \frac{1}{{\sqrt 2  \cdot \sqrt \pi  }} \cdot \frac{{\sqrt \pi  }}{{\frac{1}{{\sqrt 2 }} \cdot \sqrt {1 - 2it} }} = \frac{1}{{\sqrt {1 - 2it} }}$$.
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