Finding the Laplace Transform of f(t)=t using the definition

[mwaso]

Homework Statement

using the definition (not the table), find the Laplace transform of f(t)=t

Homework Equations

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

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?

 

[Dick]

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

 

[HallsofIvy]

Homework Statement

using the definition (not the table), find the Laplace transform of f(t)=t

Homework Equations

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

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?