Convergence of this Laplace transformation

In summary, the conversation discusses the Laplace transformation of a given function and its extension through analytic continuation. It also mentions the importance of considering the function's domain and its relation to the Laplace transform.
  • #1
lukka98
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1
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
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.
 
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  • #3
pasmith said:
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
lukka98 said:
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]
 
  • Informative
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1. What is the Laplace transformation?

The Laplace transformation is a mathematical operation that transforms a function of time into a function of frequency. It is often used in engineering and physics to solve differential equations and analyze systems.

2. How does the Laplace transformation converge?

The Laplace transformation converges when the integral used to calculate it exists. This means that the function being transformed must have certain properties, such as being bounded and having a finite number of discontinuities.

3. Why is convergence important in Laplace transformation?

Convergence is important because it ensures that the Laplace transformation is well-defined and can be used to accurately analyze a system. If the transformation does not converge, it may lead to incorrect results or solutions.

4. What happens if the Laplace transformation does not converge?

If the Laplace transformation does not converge, it means that the function being transformed does not meet the necessary criteria for the integral to exist. In this case, the Laplace transformation cannot be used to analyze the system and alternative methods must be used.

5. How can the convergence of a Laplace transformation be tested?

The convergence of a Laplace transformation can be tested using various methods, such as the Dirichlet test, Abel's test, or the Cauchy-Hadamard theorem. These tests examine the properties of the function being transformed to determine if the integral exists and the transformation converges.

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