Conditions of transformations

Jhenrique

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?

Related Differential Equations News on Phys.org

Mark44

Mentor
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."
, 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?

• 1 person

Jhenrique

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)?

Mark44

Mentor
As far as I know, there aren't any conditions on Im(s).

• 1 person

Mark44

Mentor
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

Mentor
As far as I know, there aren't any conditions on Im(s).
Jhenrique said:
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.

• 1 person

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving