Answer
$1$; $p(t)$ satisfies the given condition.
Work Step by Step
Let us consider for any $r \gt 0$:
$I=\int_{0}^{\infty} (\dfrac{1}{r}) e^{-t/r} \ dx \\=(\dfrac{1}{r}) \lim\limits_{a \to \infty} \int_0^a e^{-t/r} \ dt \\=\dfrac{1}{r} \lim\limits_{a \to \infty} (-r) [e^{t/r}]_0^a\\=\\=(-1)(-1)\\=1$
This implies that $p(t)$ satisfies the given condition.
