• Support PF! Buy your school textbooks, materials and every day products Here!

Laplace by definition

  • Thread starter mwaso
  • Start date
11
0
1. Homework Statement
using the definition (not the table), find the Laplace transform of f(t)=t

2. Homework Equations
definition of Laplace... the integral from 0 to infinity of [(e^-st)*f(t)]dt

3. The Attempt at a Solution
first I took the integral of [(e^-st)t]dt by parts and got (e^-st)(t^2)/2. I'm reasonably sure this is right cause I checked it with my TI89 :) then to evaluate it from 0 to infinity, you should be able to substitute b for infinity and take the limit as b approaches infinity. So I get the limit as b approaches infinity of (e^-sb)(b^2)/2 - (e^0)(0)/2. That second part obviously goes to 0, so we can ignore it. The first part goes to 0*infinity over 2 which isn't very helpful. However, if we move the e^-sb to the denominator, then we have the indeterminate form infinity/infinity and can utilize lohpital's (however you spell that) rule. Taking the derivative of the top and bottom yields 2b/(2se^(sb)) which is still indeterminate. Taking the derivative again gives 2/(2(s^2)*e^(sb)). This goes to zero (infinity in the denominator). However, I happen to know from my handy table of Laplace transforms that the answer I'm looking for is 1/s^2. where did I go wrong?
 

Answers and Replies

Dick
Science Advisor
Homework Helper
26,258
618
I think your TI89 is trying to trick you. Because the indefinite integral is definitely not correct.
 
HallsofIvy
Science Advisor
Homework Helper
41,738
899
1. Homework Statement
using the definition (not the table), find the Laplace transform of f(t)=t

2. Homework Equations
definition of Laplace... the integral from 0 to infinity of [(e^-st)*f(t)]dt

3. The Attempt at a Solution
first I took the integral of [(e^-st)t]dt by parts and got (e^-st)(t^2)/2. I'm reasonably sure this is right cause I checked it with my TI89 :)
No, it's wrong. Do it again, more carefully. If you still don't get the right answer, show us exactly how you integrated.

then to evaluate it from 0 to infinity, you should be able to substitute b for infinity and take the limit as b approaches infinity. So I get the limit as b approaches infinity of (e^-sb)(b^2)/2 - (e^0)(0)/2. That second part obviously goes to 0, so we can ignore it. The first part goes to 0*infinity over 2 which isn't very helpful. However, if we move the e^-sb to the denominator, then we have the indeterminate form infinity/infinity and can utilize lohpital's (however you spell that) rule. Taking the derivative of the top and bottom yields 2b/(2se^(sb)) which is still indeterminate. Taking the derivative again gives 2/(2(s^2)*e^(sb)). This goes to zero (infinity in the denominator). However, I happen to know from my handy table of Laplace transforms that the answer I'm looking for is 1/s^2. where did I go wrong?
 

Related Threads for: Laplace by definition

  • Last Post
Replies
3
Views
868
  • Last Post
Replies
7
Views
618
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
14
Views
707
  • Last Post
Replies
0
Views
742
  • Last Post
Replies
4
Views
626
Replies
1
Views
684
Replies
4
Views
7K
Top