I am having some problem to understand Convergence regions. Where is the error of following thought? Thank you in advance for some light!
I want to calculate Laplace Transform of $e^{-at}$ where $a\in\mathbb{R}^+$, what I know that is $\dfrac{1}{s+a}$ for $\text{Re}(s)>-a$.
Writing exponential in series:
$$e^{-at}=\sum_{k=0}^\infty \dfrac{a^k}{k!}.$$
So,
$$\int_0^\infty e^{-st}e^{-at} dt=\sum_{k=0}^\infty \dfrac{(-a)^k}{k!}\int_0^\infty t^ke^{-st}dt.$$
But $$\Gamma(k+1)=k!=\int_0^\infty t^ke^{-t} dt,$$ with $x>0.$
Then, for $\text{Re}(s)>0$ (*) and doing $t\to st$,
$$k!=s^{k+1}\int_0^\infty t^ke^{-st} dt.$$
So,
$$\int_0^\infty e^{-st}e^{-at} dt=\sum_{k=0}^\infty \dfrac{(-a)^k}{k!}\dfrac{k!}{s^{k+1}}=\dfrac{1}{s}\sum_{k=0}^\infty \bigg(-\dfrac{a}{s}\bigg)^k.$$
But this converges iff $\bigg\vert-\dfrac{a}{s}\bigg\vert<1$, ie, $\text{Re}(s)>a$.
Why did I not get $\text{Re}(s)>0$, since I've considered $\text{Re}(s)>0$ in (*) and the true convergence region is $\text{Re}(s)>-a$?
For instante, let be $a=1$ and $s=0.5$. So, the geometric series really diverges! But the Laplace Transform converges… So, what is the fail on the "iff"…?
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EDIT: Gary said on comments: "You can appeal to analytic continuation to extend the result to a larger domain."
Although this was not the main objective of the question that day, now if you could post an answer or an example indication of how to use analytic continuation in this specific case, I would be very grateful.
Best Answer
You obtained the result for $s>a$, say. The function $s↦1/(s+a)$ is analytic for $\Re s>−a$, and the function $$s↦∫_0^{+∞} e^{−st}e^{−at} dt$$ is also analytic for $\Re s>−a$. You showed that they agree on $s>a$. Since $\left\{ {s:s > a} \right\}$ is a subdomain of $\left\{ {s:\Re s > -a} \right\}$, by the principle of analytic continuation (or uniqueness theorem of analytic functions) they must agree on the larger domain $\Re s>−a$ as well.