= Solution
For real $p>0$, substitute $t=px$ in the <Laplace transform>:
$$
\int_0^\infty x^{a-1}e^{-px}\,dx=p^{-a}\int_0^\infty t^{a-1}e^{-t}\,dt=\boxed{\Gamma(a)p^{-a}}.
$$
The <Gamma function> integral is finite because $a>0$. For complex $p$, absolute convergence holds exactly when $\operatorname{Re}p>0$: the modulus of the integrand is $x^{a-1}e^{-(\operatorname{Re}p)x}$. On this half-plane the same answer holds with the branch $p^{-a}=\exp(-a\operatorname{Log}p)$ whose logarithm is real on the positive axis, by holomorphic continuation of the two sides.
If “exists” permits conditionally convergent ordinary improper integrals, there are additional boundary points when $0<a<1$: for $p=i\omega\ne0$, the <Dirichlet test> gives convergence, with value $\Gamma(a)|\omega|^{-a}e^{-i\operatorname{sgn}(\omega)\pi a/2}$. For $a\geq1$ these boundary integrals do not converge in that sense. At $p=0$, and for $\operatorname{Re}p<0$, the integral diverges. Thus $\operatorname{Re}p>0$ is the <region of absolute convergence of a Laplace transform>, with the stated conditional boundary qualification.
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