Why does Laplace(t) not diverge ?

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In summary, the question is about why certain functions do not have a Laplace transform. This is because the exponential term in the Laplace transform goes to zero as the variable "s" goes to infinity, making it dominant over any polynomial term. This is necessary for the Laplace transform to exist. Functions that do not have a Laplace transform include those with singular points or those with exponential terms that go to zero faster than any polynomial.
  • #1
kkirtac
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hello , anybody can explain me why Laplace(t) does not diverge ? when we perform integration by parts, the "uv" part contains the "t" variable and when we place the infinity bound here, t should cause the "uv" part to diverge as the exponential goes to zero.. so why doesn't it diverge? thanks in advance
 
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  • #2
the question is that if [tex] f(t)=O(e^{at}) [/tex] or what is the same..

[tex] f(t)e^{-at}) [/tex] for and a>0 then the F(s) Laplace transform exist for every s>a if you try to apply the Laplace transform of [tex] exp(ax^{2} [/tex] for a>0 you will say that this does not exist since for big x the quadratic term is dominant..another example of a function not having a Laplace trasnform is f(t)=1/(t-1) due to the singular point at t=1.
 
  • #3
Karlisbad said:
the question is that if [tex] f(t)=O(e^{at}) [/tex] or what is the same..

[tex] f(t)e^{-at}) [/tex] for and a>0 then the F(s) Laplace transform exist for every s>a if you try to apply the Laplace transform of [tex] exp(ax^{2} [/tex] for a>0 you will say that this does not exist since for big x the quadratic term is dominant..another example of a function not having a Laplace trasnform is f(t)=1/(t-1) due to the singular point at t=1.


thank you for your interest Karlisbad , i need a better explanation, can you or someone else explain it a bit better so that i can understand well.thank you.
 
  • #4
The point is that it still has e-st in it. That exponential goes to 0 as s goes to infinity and "dominates" any polynomial. That is, for any polynomial P(s), [itex]lim_{s\rightarrow \infty}P(s)e^{-st}= 0[/itex].

In fact the Laplace transform is only defined for functions for which that is true.
 

Related to Why does Laplace(t) not diverge ?

1. Why does Laplace(t) not diverge at certain values of t?

Laplace(t) is a mathematical function that describes the behavior of a system over time. It is defined as the integral of a function over a certain time period. At certain values of t, the function being integrated may approach infinity, causing the integral to diverge. However, in order for Laplace(t) to not diverge, the function being integrated must approach zero faster than t approaches infinity.

2. What are the consequences of Laplace(t) not diverging?

The consequences of Laplace(t) not diverging are that it allows for the use of Laplace transforms in mathematical models and systems analysis. This allows for easier analysis of complex systems and differential equations, making it a valuable tool in many scientific and engineering fields.

3. Can Laplace(t) ever diverge?

Yes, Laplace(t) can diverge at certain values of t, depending on the function being integrated. If the function approaches infinity faster than t approaches infinity, the integral will diverge.

4. What is the significance of Laplace(t) not diverging in the study of systems and control?

The fact that Laplace(t) does not diverge at certain values of t is essential in the study of systems and control. It allows for the use of Laplace transforms to analyze and control systems that would otherwise be too complex to solve using traditional methods. It also allows for the use of feedback and control systems to stabilize and optimize the behavior of complex systems.

5. Is Laplace(t) not diverging a universal property of Laplace transforms?

Yes, Laplace(t) not diverging is a universal property of Laplace transforms. This is because the Laplace transform is defined as the integral of a function over a certain time period, and in order for the integral to converge, the function being integrated must approach zero faster than t approaches infinity. Therefore, Laplace transforms will always converge as long as the function being integrated satisfies this condition.

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