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X=^-e^{t} integral

  1. Jul 21, 2005 #1
    let be the integral:

    [tex]\int_0^{\infty}F(x)dx [/tex] with [tex]x\rightarrow\infty F(x)\rightarrow0 [/tex] then we make the change of variable x=-e^{t} then the new integral would become [tex]\int_{-\infty}^{\infty+i\pi}F(-e^t)e^{t}dt[/tex] my question is if we can ignore the integral from [tex] (\infty,\infty+i\pi) [/tex] so we have only the integral [tex]\int_{-\infty}^{\infty}F(-e^t)e^{t}dt [/tex] as for big value the F(x) tends to 0
    Last edited: Jul 21, 2005
  2. jcsd
  3. Jul 21, 2005 #2


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    You cannot make this substitution. Note that x > 0, and et is also always greater than 0, so how could x = -et ever be true?
  4. Jul 21, 2005 #3
    why not? the same would happen with x=-1/(t+1) x>0 but t<0
    Last edited: Jul 21, 2005
  5. Jul 21, 2005 #4


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    How can a negative quantity equal a positive quantity?
  6. Jul 21, 2005 #5
    i think that you lost a negative sign when making your substitution.

    the example you give in post #3 is correct for x>0 t<-1

    so it sounds like the unresolved question is your bounds on the integral. i'm not sure how to handle this, but keep two things in mind:

    1) remember that you are dealing with a formal limit

    [tex]\lim_{a \rightarrow -\infty} \int_{a}^{i\pi} f(x) dx + \lim_{b \rightarrow \infty} \int_{i\pi}^{b} f(x) dx[/tex]

    2) you are integrating over [tex]\mathbb{C}[/tex] so you need to get a book on complex integration and see what to do. i suspect that solving the above and integrating as usual will suffice.
  7. Jul 22, 2005 #6
    Another question let,s suppose we have the integral:

    [tex]\int_{-\infty}^{\infty}F(x)dx [/tex] and make the change of variable x=t+ai with i
    =sqrt(-1) then what would be the new limits?..thanks...
  8. Jul 22, 2005 #7


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    I haven't taken any courses on complex analysis, so maybe I'm unfamiliar with the notation. Over what region are you integrating when you take:

    [tex]\int _{-\infty} ^{\infty} F(x)dx[/tex]

    To me, that suggests that you're integrating over all the real numbers, i.e. x ranges over the reals. If this is the case, then you again run into the problem of equating x with t + ai because unless t = b - ai for some real b, then x will be never have a real part, and t+ai sometimes will, so the two cannot be equated.
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