Inverse Laplace transform. Bromwitch integral

  • #1
587
10
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:

Answers and Replies

  • #2
18,243
7,873
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?
 
  • #3
jasonRF
Science Advisor
Gold Member
1,346
402
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
 
  • #4
587
10
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.
 

Related Threads on Inverse Laplace transform. Bromwitch integral

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
3
Views
3K
Replies
2
Views
423
Replies
2
Views
7K
Top