 Solution

For real $p>0$, substitute $t=px$ in the Laplace transform:The Gamma function integral is finite because $a>0$. For complex $p$, absolute convergence holds exactly when $\mathrm{Re}p>0$: the modulus of the integrand is $x^{a-1}{e}^{-(\mathrm{Re}p)x}$. On this half-plane the same answer holds with the branch $p^{-a}=\exp (-a\mathrm{Log}p)$ 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 $p=i\omega \ne 0$, the Dirichlet test gives convergence, with value $\Gamma (a)|\omega {|}^{-a}{e}^{-i\mathrm{sgn}(\omega )\pi a/2}$. For $a\ge 1$ these boundary integrals do not converge in that sense. At $p=0$, and for $\mathrm{Re}p<0$, the integral diverges. Thus $\mathrm{Re}p>0$ is the region of absolute convergence of a Laplace transform, with the stated conditional boundary qualification.