Convergence of this Laplace transformation

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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
 

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  • #2
pasmith
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The integral [tex]\int_0^\infty \sin t\,e^{-(1+s)t}\,dt[/tex] converges only when [itex]\operatorname{Re}(1 + s) > 0[/itex]. [itex]s = -1[/itex] is not in this region.

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

LaTeX is enabled; but it often doesn't render when you make your first post in a forum.
 
  • #3
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The integral [tex]\int_0^\infty \sin t\,e^{-(1+s)t}\,dt[/tex] converges only when [itex]\operatorname{Re}(1 + s) > 0[/itex]. [itex]s = -1[/itex] is not in this region.

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

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.
 
  • #4
pasmith
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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 [itex]\mathbb{C}[/itex] but are then extended to a larger subset of [itex]\mathbb{C}[/itex] by analytic continuation. Laplace transforms often also fall into this category.

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

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