# Laplace by definition

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?

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Dick
Homework Helper
I think your TI89 is trying to trick you. Because the indefinite integral is definitely not correct.

HallsofIvy
Homework Helper
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?