Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse Laplace transform. Bromwitch integral

  1. Nov 19, 2014 #1
    Inverse Laplace transform
    [tex]\mathcal{L}^{-1}[F(p)]=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}F(s)e^{st}dp=f(t)[/tex]

    Question if we integrate along a straight line in complex plane where axis are [tex]Re(p)[/tex], [tex]Im(p)[/tex], why we integrate from [tex]c-i \ínfty[/tex] to [tex]c+\infty[/tex]? So my question is, because [tex]Im(p)[/tex] are also real numbers why we integrate from [tex]c-i\infty[/tex] to [tex]c+i\infty[/tex].
     
    Last edited: Nov 19, 2014
  2. jcsd
  3. Nov 24, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 24, 2014 #3

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    I'm guessing that you meant that the integral is with respect to [itex]s[/itex], not [itex]p[/itex]. Anyway, for the traditional Laplace transform,
    F(s) = \int^{\infty}_{0} dt e^{-s t} f(t)

    [itex]F(s)[/itex] is only analytic for some right half plane (that is [itex]\Re(s)>s_0[/itex]). So for the inverse transform you pick a [itex]c>s_0[/itex] for your path.

    Does that help?

    jason
     
  5. Nov 25, 2014 #4
    Yes. Mistake. I understand that. I am not very good in complex analysis. My question is if [tex[Im(s)[/tex] are real numbers why I integrate from [tex]c-i\infty [/tex] to [tex]c+i\infty [/tex]? Why I have this [tex]i[/tex]? Thanks for the answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inverse Laplace transform. Bromwitch integral
Loading...