Laplace Transforms for improper integrals?

  • #1
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Homework Statement



I (came up with)/(heard about) a way of using Laplace transforms that I didn't think about before. The problem is that it doesn't work for some reason.

Look at following integral:

[tex]I = \int_{0}^{\infty }sin(t)dt[/tex]

Say that you had no idea how to integrate something. Then what you could conceivably do is consider the following function:

[tex]F(s)=\int_{0}^{\infty }e^{-st}sin(t)dt[/tex]

This converges for [itex]s > 0[/itex]. But then this is just the LT of [itex]sin(t)[/itex] (which you know), so:

[tex]F(s)=\frac{1}{s^2+1}[/tex]

Now, if [itex]s = 0[/itex], then [itex]F(0) = I[/itex], which would also mean that [itex]I = 1[/itex]. But you can't just sub in 0 because you need [itex]s[/itex] to be greater than 0. So instead take the limit as [itex]s[/itex] goes to 0, which gives the same thing. But we know that [itex]I[/itex] doesn't converge because [itex]sin(t)[/itex] oscillates...so what's going on? On a related note, this method works for some other functions like [itex]e^{-t}[/itex] ; the LT gives [itex]\frac{1}{s+1}[/itex] and then [itex]F(0) = 1[/itex].

Homework Equations



Laplace Transform

The Attempt at a Solution



Head-scratching, mainly.
 

Answers and Replies

  • #2
Dick
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I'm not sure what you are expecting here. Sure, the integral of sin(x) does not converge. The integral of sin(t)*e^(-s*t) does for s>0. And it does have a limit, again sure. What you've done is 'regularize' an integral by multiplying by e^(-s*t) to force it to converge. But the limit of that as s->0 doesn't make the original integral converge magically.
 
  • #3
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But the limit of that as s->0 doesn't make the original integral converge magically.

So does that mean this (?):

[tex] \lim_{s\to0}\int_{0}^{\infty }e^{-st}f(t)dt \neq \int_{0}^{\infty }f(t)dt [/tex]

But how can that be true? Certainly it must be true for proper integrals.
 
  • #4
Dick
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So does that mean this (?):

[tex] \lim_{s\to0}\int_{0}^{\infty }e^{-st}f(t)dt \neq \int_{0}^{\infty }f(t)dt [/tex]

But how can that be true? Certainly it must be true for proper integrals.

e^(-s*t) is less than 1 for s=>0. So yes, for any convergent integral, I think it works. But applying that to a divergent integral is a different story.
 
  • #5
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Okay, thanks! By the way, I realized what I was expecting from the LT. The limit exists - I would expect it to not exist. Why should the limit exist if the integral diverges?

If the LT of [itex]sin(t)[/itex] was something like [itex]\frac{1}{s}[/itex], then the limit as [itex]s[/itex] goes to zero doesn't exist. That would make sense, because the integral also essentially doesn't exist.
 
  • #6
Dick
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Okay, thanks! By the way, I realized what I was expecting from the LT. The limit exists - I would expect it to not exist. Why should the limit exist if the integral diverges?

If the LT of [itex]sin(t)[/itex] was something like [itex]\frac{1}{s}[/itex], then the limit as [itex]s[/itex] goes to zero doesn't exist. That would make sense, because the integral also essentially doesn't exist.

Ok, take that example. The LT of f(t)=1 is 1/s. That does diverge like the integral of 1 as s->0 because e^(-st) approaches 1 as s->0. So the limit of the LT would have to diverge as s->0. On the other hand sin(t) oscillates and the LT can average out the oscillation. Take cos(t). The limit there is 0 rather than 1. But there's not that much difference between the behavior of sin(t) and cos(t) as integrals.
 
  • #7
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Okay, I can live with that. Thanks again :smile:
 

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