How to find pdf of $Y$ given $Y=aX^2$ with $X \sim N(0,\sigma^2)$ using characteristics function ?
I only know that $\phi_X(t) = \exp(-0.5\sigma^2t^2)$ and $\phi_{aX}(t) = \phi_X(at)$ and stuck to calculate $$\phi_Y(t)=\int_{-\infty}^\infty \exp(itx^2)f_X(x) dx$$ with $f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{x^2}{2\sigma^2})$
Note $Y = a\sigma^2Z^2$, where $Z$ is $N(0, 1)$. So it suffices to compute the CHF $\phi$ of $Z^2$. We have \begin{align} \phi(t) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp(-\frac{x^2}{2} + itx^2)\,dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp(-(\frac{1}{2} - it)x^2)\,dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp(-\frac{x^2}{2(1 - 2it)^{-1}})\,dx \\ &= \frac{1}{\sqrt{2\pi}}\sqrt{2\pi}(1 - 2it)^{-1/2} \\ &= (1 - 2it)^{-1/2}. \end{align} In the fourth equality, I used the identity $$\int_{-\infty}^{\infty}\exp(-\frac{x^2}{2\sigma})\,dx = \sqrt{2\pi\sigma},$$ valid for $\Re(\sigma) > 0$. To prove this, note that the identity is clearly true for $\sigma > 0$, and that both sides of the equality define holomorphic functions of $\sigma$ for $\Re(\sigma) > 0$ (differentiate under the integral sign to see that the LHS is holomorphic). So by the identity theorem, both sides are equal for all $\sigma$ with $\Re(\sigma) > 0$.
You can recognize $\phi$ is the characteristic function of the $\text{Gamma}(\frac{1}{2}, \frac{1}{2})$ distribution. Hence $Z^2$ has the $\text{Gamma}(\frac{1}{2}, \frac{1}{2})$ density.