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Integration in Laplace Transform

  1. Jan 28, 2015 #1
    Hello everyone, I have a question about integrating in Laplace Transform. For example, if I have:
    [tex]f(t)=e^{i.t}[/tex]
    I have to solve this equation:
    [tex]\int_{0}^{\infty}e^{i.t}.e^{-s.t}dt[/tex]

    If I do like this, it's very simple:
    [tex]\int_{0}^{\infty}e^{i.t}.e^{-s.t}dt=\int_{0}^{\infty}e^{-t.(s-i)}dt=\frac{-1}{s-i}.(0-1)=\frac{1}{s-i}[/tex]

    But, if I do like this, I can't solve it:
    [tex]\int_{0}^{\infty}e^{i.t}.e^{-s.t}dt=\int_{0}^{\infty}e^{t.(i-s)}dt=\frac{1}{i-s}.(?? - 1)= ?? [/tex]

    Where it says '??', I don't know what to write.

    Any help? Thanks!
     
  2. jcsd
  3. Jan 29, 2015 #2
    You're technically dealing with a complex variable [itex]s[/itex] and an improper integral. So what you really want to calculate is
    [tex]\lim_{n \rightarrow \infty}\int_{0}^{n}e^{it}e^{-(a+b i)t}dt = \lim_{n \rightarrow \infty}\int_{0}^{n}e^{(-a+(1-b)i)t}dt = \lim_{n \rightarrow \infty} \frac{1}{-a+(1-b)i} \left ( e^{(-a+(1-b)i)n} - 1 \right) .[/tex]
     
  4. Jan 29, 2015 #3
    The Laplace transform here is defined for ##s>0##. Integrating you get $$\int_0^\infty e^{t(i-s)} dt = \frac{e^{t(i-s)}}{i-s} \Bigm|_{t=0}^{t=\infty} = \frac1{i-s} (0-1)= \frac1{s-i} .$$ Here by the value at ##t=\infty## you should understand the limit as ##t\to\infty##, and this limit is ##0## if ##s>0##. You can also consider complex ##s##, the same computation works and the limit as ##t\to\infty## is still zero if ##\operatorname{Re}s>0##. In your computations you skipped on step (did not write the antiderivative), maybe that was the source of your confusion.
     
  5. Jan 30, 2015 #4
    Thanks for your answers!
    I can see now that [tex]e^{t.(i-s)}[/tex], with [tex]t->\infty[/tex] is equal to 0, but why?
    I can separate that exponential in 2 parts: real and immaginary. In the real part, of course I have [tex]e^{-t.s}=0[/tex] but in the immaginary part I have: [tex]e^{i.t}[/tex] and that's not 0.
    So, I have [tex]0.\infty[/tex], with the infinity in the immaginary. Isn't that strange? Why do you say that is 0?
     
  6. Jan 30, 2015 #5
    For real ##t## we have ##|e^{it}|=1##, so we have something bounded (##e^{it}##) times something with limit ##0##. The resulting limit is then ##0##.
     
  7. Jan 30, 2015 #6
    Oh, that's true!!!!!!!
    Thanksssssssssssssssssssss


    Sorry, Euler, I forgot about you haha
     
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