I just started doing complex analysis and came across this notion of Analytic Continuation. Now I was thinking whether it is possible to analytically continue Laplace Transforms as well.
What I mean is this:
$$\mathcal{L}(\sin(at)) = \int_0^{\infty} e^{-st}\sin(at) \, dt $$
is only valid for $s>0$. BUT
$$ \mathcal{L}(\sin(at)) = \frac{a}{s^2 + a^2}$$
is valid unless $s \neq ia$.
So is it possible to use the above expression to find out the Laplace Transform for any other values for $s$ ?
Thanks!
Best Answer
There's a couple of things going on here. Let $f(x)$ be an exponentially bounded continuous function ($\lvert f(x) \rvert<Ke^{ax}$ uniformly for some constants $K>0$ and $a$) on $[0,\infty)$. Then $$F(s) = \int_0^{\infty} f(x) e^{-xs} \, dx $$ is an analytic function of $s$ for $\Re(s)>a$ (the integral of the derivative converges, for example). It is possible that there is an analytic function $G(s)$, equal to $F(s)$ for $\Re(s)>a$, but analytic on a larger region, although I doubt this is true without specifying more conditions on $f$: $F$ might have a natural boundary on this line. This is one reason that the inversion formula demands that the contour pass through the region where we know $F$ is analytic. Even if it can be extended, the singularities may be more complicated than points: for example the Laplace transform of $x^{1/2}$ is $\sqrt{\pi} s^{-3/2}$, which has a branch point at the origin, so can only be analytically continued to a cut plane.
But if such an extension can be made, it is often very useful for carrying out the inversion: if there are only poles, one can just find the residues. The power of the extension is that in combination with Cauchy's theorem, we can shift the integration contour about to make the inversion integral easy to calculate or estimate.