Let $X\sim N(\mu,\sigma^2)$ be independent of $Y \sim \text{Chisquared}(k)$.
I seek the pdf of the product $Z = X Y$
$X$ has pdf:
$f_X(x)={1\over\sigma\sqrt{2\pi}}e^{-{1\over2}({x-\mu\over\sigma})^2}$
$Y$ has pdf over the positive real line: $f_Y(y)={y^{(k/2)-1}e^{-y/2}\over{2^{k/2}\Gamma({k\over2})}}$
One way to find the solution is to use Rohatgi's well known result (1976,p.141) if $f_{XY}(x,y)$ be the joint pdf of continuous RV's $X$ and $Y$, the pdf of $Z$ is $f_Z(z) = \int_{-\infty}^{\infty}{{1\over|y|}f_{XY}({z\over y},y)dy} $
Since, $X$ and $Y$ are independent $f_{XY}(x,y)=f_X(x)f_Y(y)$ $f_Z(z) = \int_{-\infty}^{\infty}{{1\over|y|}f_X({z\over y})f_{Y}(y)dy} $ $f_Z(z) = {1\over\sigma_x\sqrt{2\pi}}{1\over{2^{k/2}\Gamma({k\over2})}}\int_{0}^{\infty}{{1\over|y|}e^{-{1\over2}({{z\over y}-\mu_x\over\sigma_x})^2} {y^{(k/2)-1}e^{-y/2}}dy} $ ... where we face the problem of solving the integral $\int_{0}^{\infty}{e^{-{1\over2}({{z\over y}-\mu_x\over\sigma_x})^2} {y^{(k/2)-2}e^{-y/2}}dy}$.
Can anyone help me with this problem?