Does the Laplace transform of a probability distribution function (pdf) also integrates to 1?

Question

I have a pdf $p(t)$ $$\int_{0}^{\infty}p(t)dt=1$$ now I take its laplace transform as: $P(s)$ should $$\int_{0}^{\infty}P(s)ds=1?$$ Intuitively, I think it should integrate to $1$.

Answer 1

No, $\int_0^{\infty} P(s)\, ds$ can be $\infty$. For example, take $p(t)=\frac 1 {2\sqrt t}$ for $0<t<1$ and $0$ for all other $t$.

We have $\int_0^{\infty} P(s)\, ds=\int_0^{1} \int_0^{\infty}e^{-st} \frac 1 {2\sqrt t} \, ds\, dt=\frac 1 2\int_0^{1} t^{-3/2}\, dt=\infty$.