Laplace transform of a piecewise function

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Homework Help Overview

The discussion revolves around finding the Laplace transform of a piecewise function defined as f(t) = e^t for 0≤t<1 and f(t) = 0 for t≥1. Participants are exploring the integration process involved in computing the Laplace transform.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the integral for the Laplace transform, with some attempting to clarify the correct form of the integrand. Questions arise regarding the treatment of the piecewise nature of the function and the implications of the integral from 1 to infinity.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's attempts. There is recognition of mistakes made in the integration process, and some participants express a need for clarification on the integrand. No consensus has been reached yet, but there is a collaborative effort to address misunderstandings.

Contextual Notes

Participants are navigating potential confusion between the Laplace transform and other transformations, as well as addressing errors in their initial interpretations of the function's definition.

Feodalherren
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Homework Statement



f(t) = e^t when 0≤t<1
and 0 when t≥1

Homework Equations


Laplace transformations

The Attempt at a Solution



so the Laplace integral becomesfrom 0 to 1 ∫e^(st^2)dt + 0

how do I integrate this?
 
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Clarification:

Laplace transform or Lagrange Transform...?
 
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My bad, Laplace :).
 
Alright, let's do this:

From 0 to 1 we have one function, and from 1 onward we have another. Split up our integral as so:

$$\int_0^1 e^{-st} e^{t}dt + \int_{1}^{\infty} e^{-st}(0)dt \implies \int_0^1 e^{t(1-s)}dt$$
 
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Wait a second, doesn't the part that goes from 1 to +infinity get canceled out because the integral becomes

∫ e^(-st) (0) dt = 0

?
 
Feodalherren said:
Wait a second, doesn't the part that goes from 1 to +infinity get canceled out because the integral becomes

∫ e^(-st) (0) dt = 0

?

Excuse my reading comprehension, I thought it said f(t) = 1. Corrected.
 
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Ah now I see what I did wrong! DUH! Such a stupid mistake.

Thank you sir!
 
Feodalherren said:

Homework Statement



f(t) = e^t when 0≤t<1
and 0 when t≥1

Homework Equations


Laplace transformations

The Attempt at a Solution



so the Laplace integral becomesfrom 0 to 1 ∫e^(st^2)dt + 0

how do I integrate this?
How did you get an integrand of est2 ?

Remember, ex ⋅ ey = e(x + y), not exy :wink:
 
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I was just asking myself the same thing. I think I need to take a break. I've been doing math since 7.30 this morning. It's 1.30 pm now :).
 

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