TITLE: $\int_{0}^{\infty}e^{-st}h(t)dt=0 \Rightarrow h(t)=0.$
QUESTION [6 upvotes]: Suppose $h(t)$ is continuous function and $\int_{0}^{\infty}e^{-st}h(t)dt=0 ~\forall~ s>s_{0}$, then prove that $h(t)=0$.
I know "if a function is continuous, non-negative or non-positive, and its integration is zero, then function must be zero", which is intuitively clear.   
But here asked question is beyond my knowledge. Here, exponential function is doing some miracle, but how?
Would you like to help me?

REPLY [2 votes]: The assumption is equivalent to
$$
\int_0^\infty e^{-sx}\left(e^{-s_0x}h(x)\right)\,\mathrm{d}x=0\tag{1}
$$
for all $s\gt0$
Since
$$
\begin{align}
\int_0^\infty\left(e^{-x}-e^{-2x}\right)^n\,\mathrm{d}x
&=\int_0^\infty e^{-(n-1)x}\left(1-e^{-x}\right)^n\,\mathrm{d}e^{-x}\\
&=\int_0^1t^{n-1}(1-t)^n\,\mathrm{d}t\\
&=\frac{n!(n-1)!}{(2n)!}\tag{2}
\end{align}
$$
define
$$
f_n(x)=n\binom{2n}{n}\left(e^{-x}-e^{-2x}\right)^n\tag{3}
$$
Then
$$
\int_0^\infty f_n(x)\,\mathrm{d}x=1\tag{4}
$$
Furthermore, for $x\gt0$,
$$
0\lt4\left(e^{-x}-e^{-2x}\right)\le1\tag{5}
$$
and if $x\ne\log(2)$, then the right inequality is strict.
According to $(9)$ from this answer,
$$
\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}\tag{6}
$$
Combining $(3)$, $(5)$, and $(6)$, we get that for $x\gt0$
$$
0\lt f_n(x)\le\sqrt{\frac{n}\pi}\,\left[4\left(e^{-x}-e^{-2x}\right)\right]^n\tag{7}
$$
so that outside any neighborhood of $\log(2)$, $f_n(x)$ eventually tends monotonically to $0$.
$(4)$ and $(7)$ imply that $\frac{\log(2)}\alpha f_n\left(\frac{\log(2)}\alpha x\right)$ is an approximation of $\delta(x-\alpha)$.
According to the assumption, for any $n$ and $\alpha$,
$$
\int_0^\infty\frac{\log(2)}\alpha f_n\left(\frac{\log(2)}\alpha x\right)\left(e^{-s_0x}h(x)\right)\,\mathrm{d}x=0\tag{7}
$$
Since $h(x)$ is continuous, $(7)$ implies that $h(x)=0$ for $x\ge0$.