I'd like to verify that $\mathcal{L}[e^{-at}]=\frac{1}{s+a}$, $t\ge 0$.
So I calculate: $$\int_{0^+}^{+\infty} e^{-at} e^{-st} dt=\int_{0^+}^{+\infty} e^{-t(a+s)} dt = \frac{1}{-(a+s)} e^{-t(a+s)}\Big|_{0^+}^{+\infty}$$.
Now I have to distinguish three cases:
1) $(a+s)\gt0$: $\frac{1}{-(a+s)} e^{-t(a+s)}\Big|_{0^+}^{+\infty}=\frac{1}{a+s}$
2) $(a+s)=0$: impossible (because of the domain)
3) $(a+s)\lt0$: $\frac{1}{-(a+s)} e^{-t(a+s)}\Big|_{0^+}^{+\infty}=+\infty+\frac{1}{a+s}$
So is $\mathcal{L}[e^{-at}]=\frac{1}{s+a}$ only valid when $a+s\gt0$?
Also: since $s \in \mathbb{C}$, the whole thing has no sense (since the inequations in $\mathbb{C}$ don't make sense) ?
EDIT
If I write $s=a+ib=|s|e^{i\theta}=|s|(cos\theta+isin\theta) \in \mathbb{C}$, I get:
$$ \begin{align} \int_{0^+}^{+\infty} e^{-at} e^{-st} dt&=\int_{0^+}^{+\infty} e^{-at} e^{-|s|(cos \theta +i sin \theta)t} dt\\ &=\int_{0}^{+\infty} e^{-(a+|s|+cos \theta +i sin \theta )t}\\ &=-\frac{e^{-(a+|s|+cos \theta + i sin \theta )t}}{a+|s|+cos \theta + i sin \theta}\Big|_{0}^{+\infty} \end{align} $$
But here too, I have to see if $a+|s|+cos \theta + i sin \theta$ is positive or not. Do I?