1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    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

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    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

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: X=^-e^{t} integral
  1. V(x) to V(t) (Replies: 3)

Loading...