Derive density of $Z=XY$, $X\sim U(0,1)$ and $Y\sim\mathcal N(0,1)$. I have been stumped for a few days on this. I have two random variables $X\sim U(0,1)$ and $Y\sim\mathcal N(0,1)$, which are independent. How can I get the density of $Z = XY$?
I put $Z = XY, W = Y$ i.e. $X=Z/W, Y=W$ and achieved the Jacobian as $J={1\over|W|}$,
but I've got
$f_{Z,W}(z,w)={1\over\sqrt{2\pi}|w|}{\exp(-w^2/2)}$
and I don't know how to integrate this w.r.t $w$ and get the (marginal) density of $Z$.
Could you please help me with this problem? I tried partial integration too but it doesn't work.


I put $Z = XY, W = Y$ i.e. $X=Z/W, Y=W$ and achieved the Jacobian as $J={1\over|W|}$, but I've got $f_{Z,W}(z,w)={1\over\sqrt{2\pi}|w|}{\exp(-w^2/2)}$
and I don't know how to integrate this w.r.t $w$ and get the (marginal) density of $Z$.

That is okay, but you forgot the support.   Always note the support!    Doing so will give you the same result as @FelixMarin obtained.
$\qquad\begin{align}f_{Y,Z}(y,z)&=f_{Y,XY}(y,z) \\&= \begin{vmatrix}\dfrac{\partial z/y}{\partial z}\end{vmatrix} f_{X,Y}(z/y,y)\\&=\begin{vmatrix}\dfrac{1}{y}\end{vmatrix}\dfrac{\exp(-y^2/2)}{\surd(2\pi)}\,\mathbf 1_{0\leq z/y\leq 1}\,\mathbf 1_{y\in\Bbb R}\\&=\begin{vmatrix}\dfrac{1}{y}\end{vmatrix}\dfrac{\exp(-y^2/2)}{\surd(2\pi)}\,(\mathbf 1_{0\leq z\leq y}+\mathbf 1_{y\leq z\leq 0})\\[3ex] f_Z(z) &= \mathbf 1_{0\leq z}\int_z^\infty \dfrac{\exp(-y^2/2)}{y\surd(2\pi)}\mathrm d y-\mathbf 1_{z<0}\int_{-\infty}^z\dfrac{\exp(-y^2/2)}{y\surd(2\pi)}\mathrm d y\\&=\mathbf 1_{z\in\Bbb R}\int_{\lvert z\rvert}^\infty\dfrac{\exp(-y^2/2)}{y\surd(2\pi)}\mathrm d y\\&~~\vdots\\&=\dfrac{\Gamma(0, z^2/2)}{2\surd(2\pi)}\,\mathbf 1_{z\in\Bbb R}\end{align}$
Where the incomplete gamma function is defined as $\Gamma(a,x):=\int_x^\infty t^{a-1}\exp(-t)\,\mathrm d t$ so
$$\Gamma(0, z^2/2) = \int_{z^2/2}^\infty t^{-1}\exp(-t)\,\mathrm d t$$
