What is the inverse laplace transform of this?

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Discussion Overview

The discussion centers on finding the inverse Laplace transform of the function F(s) = 1/K^s, where K is a positive real number. The scope includes mathematical reasoning and exploration of the properties of the Laplace transform.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant proposes rewriting F(s) as exp[-s Log(K)] to facilitate comparison with the Laplace integral.
  • Another participant suggests that if f(t) is a function with a large peak around t = Log(K), it could yield the correct Laplace transform, albeit with some normalization.
  • A question is raised about whether the inverse Laplace transform is δ(t - log(K)).
  • A later reply confirms the suggestion that it is indeed δ(t - log(K)).

Areas of Agreement / Disagreement

Participants appear to agree on the conclusion that the inverse Laplace transform is δ(t - log(K)), but the reasoning leading to this conclusion remains open for further exploration.

Contextual Notes

The discussion does not clarify the normalization factor mentioned, nor does it resolve the implications of the large peak in f(t) on the overall validity of the inverse transform.

Charles49
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[tex]F(s) = \frac{1}{K^s}[/tex]

where K is a positive real.
 
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Hint:

Write this as exp[-s Log(k)]

Then compare this to the Laplace integral:

Integral of exp(-s t) f(t) dt

So, it looks like if you take f(t) to be a function that has a very large peak around t = Log(K), you'll get the correct Laplace transform up to some normalization. Now think of making this line of reasoning more precise...
 
So is it [tex]\delta(t-\log(K))[/tex]?
 
Charles49 said:
So is it [tex]\delta(t-\log(K))[/tex]?

That's right!
 

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