Results 1 to 3 of 3

Math Help - Simple Laplace transform problem

  1. #1
    Junior Member Evales's Avatar
    Joined
    Oct 2007
    Posts
    63

    Simple Laplace transform problem

    Exam study haha

    2. Consider
    f(t) = e^at for t > 0, where 'a' is a constant, then

    F
    (s) = Integral(from 0 to infinity) e^(st)e^(at)dt


    and this integral exists for
    s > a, so that

    L
    (eat) = 1/ (s-a)


    for
    s > a.


    I would have said 1/ (a-s)
    Why is it not so?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Evales View Post
    Exam study haha


    2. Consider f(t) = e^at for t > 0, where 'a' is a constant, then

    F
    (s) = Integral(from 0 to infinity) e^(st)e^(at)dt

    and this integral exists for


    s > a, so that
    L


    (eat) = 1/ (s-a)

    for s > a.


    I would have said 1/ (a-s)
    Why is it not so?

    \int_{0}^{+\infty} e^{-st} e^{at} \, dt = \int_{0}^{+\infty} e^{-(s - a)t} \, dt = ....

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328
    \int_0^\infty e^{at}e^{-st}dt= \int_0^\infty e^{(a-s)t}dt

    Let u= (a-s)t so that du= (a-s)dt and du/(a-s)= dt. Now the integral is \frac{1}{a-s}\int e^udu but we have to be careful about the limits of integration. At t= 0, u= (a-s)t= (a-s)0= 0 but if s> a, a-s< 0 so the upper limit will be -\infty.

    That is, the integral is now \frac{1}{a-s}\int_0^{-\infty} e^u du. Integrating that gives, again, an exponential which will be 0 at the upper limit. The integral is 0- \frac{1}{a-s}e^0= \frac{1}{s-a}.
    Last edited by mr fantastic; November 4th 2009 at 03:23 AM. Reason: Fixed some latex tags
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: June 1st 2011, 04:49 AM
  2. [SOLVED] Simple Inverse Laplace Transform
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: March 16th 2011, 01:20 PM
  3. Laplace Transform problem
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: April 8th 2010, 06:15 AM
  4. Replies: 1
    Last Post: January 25th 2009, 08:50 PM
  5. Replies: 7
    Last Post: December 11th 2008, 08:16 PM

Search Tags


/mathhelpforum @mathhelpforum