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Inverse Laplace transform

  1. Dec 13, 2011 #1
    [tex]f(t)=\int^{c+i\infty}_{c-i\infty}F(s)e^{st}ds[/tex]

    Why we suppose that all singularities are left from line [tex]Re(s)=c[/tex]?
     
  2. jcsd
  3. Dec 13, 2011 #2
    Here is picture of my question. Can you give me detail explanation?
     

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  4. Dec 13, 2011 #3
    We place the path of integration (by design and not by necessity), to the right of all singularities so that the value of the integral is independent of the specific value of c. You can show that by evaluating the integral over a square contour which goes around to the right and since there are no singularities there, the value of the integral is zero. Now due to the restrictions placed on functions which have laplace transforms (those of exponential order), the horizontal legs on the top and bottom of that contour can be shown to be zero which means the sum of the two vertical legs are zero which means they are equal to one another when both are going in the same direction which means the integral is independent of the value of c.
     
  5. Dec 13, 2011 #4
    I can't understand you really well without a picture. But I understand that result is for any [tex]Re(s)=c[/tex] this is correct. Fine. I don't have a problem with that. My problem is that I certainly can choose a lot of counture so that I have 3, for example, isolated singularities in the right side of [tex]Re(s)=c[/tex]. Am I right?

    My second question is:
    When I defined Laplase transform like

    [tex]F(s)=\int^{\infty}_0f(t)e^{-st}dt[/tex]

    I say that [tex]Re(s)>0[/tex] because of convergence. So [tex]c[/tex] must be positive. Right? Is there some other condition?
     
  6. Dec 13, 2011 #5
    If I had conture like in picture 2 in file

    http://www.solitaryroad.com/c916.html

    Why integral [tex]\int_{C_1}=0[/tex] when [tex]R\rightarrow \infty[/tex]? I have

    [tex]\lim_{R\to \infty}\int_{C_1}F(s)e^{st}dp=0[/tex]

    If I understand [tex]R\rightarrow \infty[/tex] is equivalent with [tex]\lim_{Re(s)\to \infty}[/tex].

    [tex]\lim_{Re(s)\to \infty}F(s)=0[/tex]

    and

    [tex]\lim_{Re(s)\to \infty}e^{pt}=\infty[/tex]

    why then

    [tex]\lim_{R\to \infty}\int_{C_1}F(s)e^{st}dp=0[/tex]?
     
  7. Dec 14, 2011 #6
    Can someone help me and answer these questions?
     
  8. Nov 20, 2012 #7
    Can you give me a explanation why all singularities are left from ##Re(s)=c##? And why we integrate over the line ##(c-i\infty,c+i\infty)##?
     
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