I Convergence of this Laplace transformation

[lukka98]

I have a f(t) that is, e^(-t) *sin(t), now I calculate the Laplace transformation, that is:
X(s) = 1 / ( 1 + ( 1 + s)^2 ) (excuse me but Latex seems not run ).
Now I imagine the plane with Re(s), Im(s) and the magnitude of X(s).

If i take Re(s) = -1 and Im(s) = 0, I believe I have X(s) = 1 ( s = -1, so from the formula X(s) = 1) and this seem correctly according to a graph that I see online.

But if I put Re(s) = -1 and Im(s) = 0 in the integral, and I calculate it directly, I should get the same result...(?) but In that case is only the integral of sin(t) from 0 to infty that is not 1 absolutely.

What is wrong?

thanks

 

[pasmith]

The integral \int_0^\infty \sin t\,e^{-(1+s)t}\,dt converges only when \operatorname{Re}(1 + s) > 0. s = -1 is not in this region.

You can, of course, analytically continue the result you get in \operatorname{Re}(1 + s) > 0 into the rest of the s-plane (except the poles at s = -1 \pm i).

LaTeX is enabled; but it often doesn't render when you make your first post in a forum.

 

[lukka98]

The integral \int_0^\infty \sin t\,e^{-(1+s)t}\,dt converges only when \operatorname{Re}(1 + s) > 0. s = -1 is not in this region.

You can, of course, analytically continue the result you get in \operatorname{Re}(1 + s) > 0 into the rest of the s-plane (except the poles at s = -1 \pm i).

LaTeX is enabled; but it often doesn't render when you make your first post in a forum.

Ok, maybe my question was misunderstood, but $X(s) = \frac{1}{1 + (1+s)^2}$, so for s = -1, I have X(s) = 1;
but $\int_{0}^{\infty} sin(t) dt$ ,that is for s = -1, is indefinite.
And also the graphics solution, looks like X(s) = 1 for s = -1.

 

[pasmith]

Ok, maybe my question was misunderstood, but $X(s) = \frac{1}{1 + (1+s)^2}$, so for s = -1, I have X(s) = 1;
but $\int_{0}^{\infty} sin(t) dt$ ,that is for s = -1, is indefinite.
And also the graphics solution, looks like X(s) = 1 for s = -1.

Yes.

There are many functions, such as the Gamma function, which are initially defined by an integral representation valid only in a subset of \mathbb{C} but are then extended to a larger subset of \mathbb{C} by analytic continuation. Laplace transforms often also fall into this category.

One can define a function F : D \to \mathbb{C}: s \mapsto \int_0^\infty f(t)e^{-st}\,dt where D \subset \mathbb{C} is the region where the integral converges. If D satisfies certain conditions then there exists a unique G, defined on as much of \mathbb{C} as possible, such that G(s) = F(s) for every s \in D.

Now F and G are distinct functions because they have different domains, and we commonly define the Laplace transform of f as F when really we mean G: Inverting the transform requires integration over a contour which is almost certainly not contained in D, and if we have a formula for F(s) then it almost certainly makes sense for values of s \notin D. But we must remember that if s \notin D then <br /> G(s) \neq \int_0^\infty f(t)e^{-st}\,dt.