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Upper limit gives zero

  1. Nov 19, 2007 #1
    I was performing a Laplace transformation problem,where I happened to face:

    {lower limit constant and upper limit infinity} exp [(i-s)t] the variable being t.

    I am not sure if the upper limit gives zero,but if I assume that the answer becomes correct.

    Can anyone please tell me why the upper limit is zero here?
  2. jcsd
  3. Nov 19, 2007 #2
    OK,I am writing it in LaTeX:

    [tex]\int^\infty_{2\pi/3}[/tex] [tex]\ e^{(i-s)t} [/tex] [tex]\ dt [/tex]
  4. Nov 19, 2007 #3
    Friends,I do not know why the exponential dd not appear.Please assume this and let me know:
  5. Nov 19, 2007 #4


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    Well, obviously, the anti-derivative is
    [tex]\int e^{(i-s)t}dt= \frac{1}{i-s}e^{(i-s)t}[/tex]
    IF s> i, then the limit as t goes to infinity will be 0. If [itex]s\le i[/itex] the integral does not exist.
  6. Nov 19, 2007 #5
    I see...Thank you.

    Can you please tell me why the integral does not exist for i<s?
  7. Nov 19, 2007 #6

    Ben Niehoff

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    Moreover, what does an expression like [itex]s \leq i[/itex] mean, considering that the complex numbers are not ordered?

    If [itex]s = \sigma + i\omega[/itex], then

    [tex]\begin{array}{rcl}\frac 1 {i - s} e^{(i -s)t} & = & \frac1 {i - \sigma - i\omega} e ^{(i- \sigma - i\omega)t}\\&&\\ &=& \frac 1 {-\sigma + i(1-\omega)} e^{i(1-\omega)t} e^{-\sigma t}\end{array}[/tex]

    which has a finite limit at infinity only if [itex]\sigma > 0[/itex], and hence [itex]\Re(s) > 0[/itex].

    Hmm...maybe that's what was meant originally, then.
  8. Nov 19, 2007 #7
    Yes,I afree.
    If you take the modulus,the ghost of exp[it] runs away and the thing goes to zero as t tends to infinity
  9. Nov 20, 2007 #8


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    My mistake. I had not realized that "i" was the imaginary unit! In that case, separate it into two parts. [itex]e^{(i-s)t}= e^{it}e^{-st}[/itex]. The [itex]e^{it}[/itex] part oscilates (it is a sin, cos combination) while e^{-st} will go to 0 as t goes to infinity because of the negative exponential. The entire product goes to 0 very quickly so the anti-derivative goes to 0.

    (The "ghost" of eit- I like that.)
    Last edited by a moderator: Nov 20, 2007
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