Integration in Laplace Transform

  • Thread starter juan.
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  • #1
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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!
 

Answers and Replies

  • #2
137
5
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]
 
  • #3
177
61
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.
 
  • #4
8
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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?
 
  • #5
177
61
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##.
 
  • #6
8
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Oh, that's true!!!!!!!
Thanksssssssssssssssssssss


Sorry, Euler, I forgot about you haha
 

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