Two step random experiment: Density of combined uniform and normal distribution imagine a random experiment, where first some number $u$ is drawn uniformly on $[c-\varepsilon,c+\varepsilon]$ for $c>0$ and $0<\varepsilon<c$. Next, a $N(u,\sigma^2)$-distributed random variable $Y$ is generated (meaning that we use the $u$ of the first step as mean and assume some known variance $\sigma^2>0$ for the normal distribution). Now, I am interested in the density of $Y$. How can one derive this density? Should I compute the joint density of $u$ and $Y$?
Thanks a lot!

Density of $Y$:
\begin{align}
f_Y(y) &= \int_{u=c-\epsilon}^{c+\epsilon} f_{Y|U}(y\mid u)f_U(u)\;du \\
&= \int_{u=c-\epsilon}^{c+\epsilon} \dfrac{1}{2\epsilon}\dfrac{1}{\sqrt{2\pi\sigma^2}}\;e^{-\dfrac{1}{2}\left(\dfrac{y-u}{\sigma}\right)^2}\;du. \\
\end{align}
Let $z=(u-y)/\sigma\;$ so that $dz=du/\sigma$. Then,
\begin{align}
f_Y(y) &= \dfrac{1}{2\epsilon}\int_{z=(c-\epsilon-y)/\sigma}^{(c+\epsilon-y)/\sigma} \dfrac{1}{\sqrt{2\pi}}\;e^{-\frac{1}{2}z^2}\;dz \\
& \\
&= \dfrac{1}{2\epsilon}\left[ \Phi\left(\dfrac{c+\epsilon-y}{\sigma}\right) - \Phi\left(\dfrac{c-\epsilon-y}{\sigma}\right)\right].
\end{align}
