Understanding what happens to exp((6-s)t) when t goes to inf

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Homework Help Overview

The discussion revolves around the behavior of the expression exp((6-s)t) as t approaches infinity, particularly in the context of Laplace Transforms. The original poster seeks to understand the implications of this limit and how it relates to the existence of Laplace Transforms for certain values of 's'.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the conditions under which the Laplace Transform exists, particularly questioning the significance of the inequality (6-s)<0. There is also discussion about the behavior of the exponential terms as t approaches infinity and the implications for the validity of the Laplace Transform.

Discussion Status

Participants are actively engaging with the problem, raising questions about the assumptions underlying the Laplace Transform and clarifying the conditions necessary for its existence. Some have pointed out the need to consider the real part of 's' in the context of convergence.

Contextual Notes

There is an acknowledgment that not all functions have Laplace transforms, and the discussion includes considerations of specific regions of 's' for which the transforms are valid. The distinction between evaluating limits and simply substituting infinity is also noted.

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Homework Statement



I want to understand what happens to exp((6-s)t) when t goes to infinity. Btw, this is part of a Laplace Transform so that's what the 's' is.

Homework Equations



f(t) = exp(6t)u(t)

Find Laplace Transform.

The Attempt at a Solution



I get to:

exp((6-s)t) / (6-s) in which i must integrate from 0 to infinity. I'm pretty sure the answer is 1 / (6-s) but I don't understand why. Breaking apart exp(6t) goes to infinity. And breaking apart exp(-st) goes to 0 right? *when t=inf
 
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It's okay if (6-s)<0.
 
For a Laplace Transform is that always true?
 
No, I just figured something out. Not all functions have Laplace transforms, the example of the exponential increase is one example of that.
 
Many functions f(t) have Laplace transforms F(s) that are valid only for certain regions of s. For example, u(t) has L.T. = 1/s, but this is valid only if s > 0 because it comes from integrating exp(-s*t) for t from 0 to +infinity.

RGV
 
So when applying t=infinity for the function exp((6-s)t) , it doesn't exist? Or does exp(6t) just get neglected?

EDIT: just saw ray's post. So must it be specified that (6-s)<0 or is it just something known with Laplace Transforms?
 
oso0690 said:
So when applying t=infinity for the function exp((6-s)t) , it doesn't exist? Or does exp(6t) just get neglected?

EDIT: just saw ray's post. So must it be specified that (6-s)<0 or is it just something known with Laplace Transforms?
No, you cannot neglect it; that is why the L.T. makes no sense for s < 6. And, of course, we never let t = infinity; we take the *limit* as t --> infinity. (That is how integrals are _defined_ over infinite intervals: as limits.) Anyway, when s > 6 the limit = 0, so the term goes away. However, we did not "neglect" it; we examined it carefully.

RGV
 
Thank you for correcting me.

So here's what I have:

1/(6-s) * (exp((6-s)t) where t is evaluated from 0 to infinity

1/(6-s) * ( exp((6-s)*inf) - exp((6-s)*0) )

For s > 6

1/(6-s) * (0 - 1)

-1/(6-s)

1/(s-6)

For s < or equal to 6, L.T. does not exist.

Is this correct?

Btw thanks to both of you for helping me.
 
Remember s is complex, so it doesn't make sense to say s>6. What you should say is Re(s)>6, where Re(s) denotes the real part of s. The half plane Re(s)>6 is called the region of convergence.
 

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