I'm confused about how to show where a Laplace transform converges.
E.g. take $f(x)=e^{ax}$ where $a\in\mathbb{C}$ is fixed. Then the Laplace transform of $f$ is
$$\bar{f}(p)=\int_0^\infty e^{ax}e^{-px} \, \mathrm dx=\int_0^\infty e^{-(p-a)x} \, \mathrm dx$$.
My notes say that this equals $\frac{1}{p-a}$, provided $\operatorname{Re}(p) > \operatorname{Re} (a)$.
How do I show this last part explicitly? This is basically saying that the integral of the exponential diverges if its argument has positive real part. How do I spell this out line by line?
$\int_0^{t} e^{-(p-a)}dx=\frac 1 {p-a} (1-e^{-(p-a)t})$. Note that $|e^{-(p-a)t}|=e^{-\Re (p-a)t}\to \infty$ if $\Re (p-a)<0$ and $|e^{-(p-a)t}|\to 0$ if $\Re (p-a)>0$. I will leave the rest to you.