Solution
= Solution
For $U\sim\operatorname{Exponential}(1)$,
$$
\bar S(t)=\int_0^\infty e^{-u\theta t}e^{-u}du
=\frac1{1+\theta t},
\qquad
\bar h(t)=\frac\theta{1+\theta t}.
$$
Again $\bar h(0)=\theta$ because $\mathbb EU=1$, but now $\bar h(t)\to0$ as $t\to\infty$.