What are the conditions for using the Laplace transform?

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Discussion Overview

The discussion centers on the conditions for using the Laplace transform, particularly focusing on the implications of the complex variable \( s \) and its real and imaginary components. Participants explore the assumptions required for the integral to converge, as well as the treatment of complex numbers in this context.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that the Laplace transform of 1 is given by the integral \( \int_{0}^{\infty} 1 \exp(-st) dt \) and that it converges under the assumption that \( s > 0 \).
  • Others clarify that the correct assumption is \( \text{Re}(s) > 0 \) and reference external sources for confirmation.
  • A participant questions the conditions on \( \text{Im}(s) \), suggesting that there may be implications for the imaginary part of \( s \).
  • Another participant states that, to their knowledge, there are no conditions on \( \text{Im}(s) \), prompting further inquiry into why this might be the case.
  • One participant discusses the definition of the Laplace transform and its convergence criteria, noting that the integral converges for some real number \( \sigma \) when \( \text{Re}(s) > \sigma \).
  • Concerns are raised about the use of \( \infty \) in arithmetic expressions, with a participant arguing that such usage is not meaningful.
  • References to various mathematical texts are made, indicating differing approaches to the treatment of complex values in the context of Laplace transforms.

Areas of Agreement / Disagreement

Participants generally disagree on the conditions related to \( \text{Im}(s) \) and the implications of using complex numbers in the Laplace transform. There is no consensus on whether conditions on \( \text{Im}(s) \) are necessary or relevant.

Contextual Notes

Limitations in the discussion include the lack of clarity on the implications of the imaginary part of \( s \) and the varying definitions of convergence in different mathematical texts. The treatment of \( \infty \) in expressions remains a point of contention.

Jhenrique
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The Laplace of 1 is:

$$\int_{0}^{\infty} 1 \exp(-st) dt = \left[ \frac{\exp(-st)}{-s} \right]_{0}^{\infty} = \frac{\exp(-s \infty) - \exp(-s 0)}{-s} = \frac{0 - 1}{-s} = \frac{1}{s}$$

It's result known, however, for this be true is assumed that s>0, because 0 = exp(-∞) = exp(-s∞). But we have a problem, s is a complex number (σ + iω), so you assume that s>0 thus you are saying that ω=0, but in laplace transform ω≠0. The most correct possible would be ##\exp(-s∞) = \exp(-(σ + iω)∞) = \exp(-σ∞ - iω∞) = \exp(-\text{sgn}(σ)∞ - i \text{sgn}(ω)∞) = \frac{\exp(-\text{sgn}(σ)∞)}{\exp(i \text{sgn}(ω)∞)}##, but it's become inpraticable... So what is correct form of approach this calculation of a simple way?
 
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Jhenrique said:
The Laplace of 1 is:

$$\int_{0}^{\infty} 1 \exp(-st) dt = \left[ \frac{\exp(-st)}{-s} \right]_{0}^{\infty} = \frac{\exp(-s \infty) - \exp(-s 0)}{-s} = \frac{0 - 1}{-s} = \frac{1}{s}$$

It's result known, however, for this be true is assumed that s>0
No, the assumption is that Re(s) > 0. See http://en.wikipedia.org/wiki/Laplace_transform, Table of selected Laplace transforms. Note the column whose heading is "Region of convergence."
Jhenrique said:
, because 0 = exp(-∞) = exp(-s∞). But we have a problem, s is a complex number (σ + iω), so you assume that s>0 thus you are saying that ω=0, but in laplace transform ω≠0. The most correct possible would be ##\exp(-s∞) = \exp(-(σ + iω)∞) = \exp(-σ∞ - iω∞) = \exp(-\text{sgn}(σ)∞ - i \text{sgn}(ω)∞) = \frac{\exp(-\text{sgn}(σ)∞)}{\exp(i \text{sgn}(ω)∞)}##, but it's become inpraticable... So what is correct form of approach this calculation of a simple way?
 
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Mark44 said:
No, the assumption is that Re(s) > 0. See http://en.wikipedia.org/wiki/Laplace_transform, Table of selected Laplace transforms. Note the column whose heading is "Region of convergence."

Humm, but s is a complex number, so what is the conditition for Im(s)?
 
As far as I know, there aren't any conditions on Im(s).
 
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Jhenrique said:
The most correct possible would be ##\exp(-s∞) = \exp(-(σ + iω)∞) = \exp(-σ∞ - iω∞) = \exp(-\text{sgn}(σ)∞ - i \text{sgn}(ω)∞) = \frac{\exp(-\text{sgn}(σ)∞)}{\exp(i \text{sgn}(ω)∞)}##
What you wrote above is pretty much meaningless. You cannot use ∞ in arithmetic expressions.
 
Mark44 said:
As far as I know, there aren't any conditions on Im(s).

Why not?
 
Mark44 said:
As far as I know, there aren't any conditions on Im(s).

Jhenrique said:
Why not?
Maybe they're not needed. I dug up my book on Advanced Engineering Mathematics, Third Edition (Erwin Kreyszig) and looked at the section where he talks about Laplace tranforms. He doesn't even mention that s is complex. To see a treatment on the Laplace transform that deals with complex values of s, I looked at my copy of Churchill's Complex Analysis, which doesn't have any topics on this tranform, and then my Jerrold Marsden "Basic Complex Analysis," which does include this topic.

The Laplace transform is usually defined as this integral (which I believe you know):
$$F(s) = \int_0^{\infty}e^{-st}f(t)dt$$

The defining theorem for this transform says that there is some real number σ for which the above integral converges if Re(s) > σ, and diverges if Re(s) < σ.

If you want to know more than that, you'll need to study complex analysis.
 
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