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Proof that x^s/s equals 1 if x > 1 where the function is integrated with limits

  1. Nov 29, 2009 #1
    can someone prove this for me using a semicircle contour as opposed to a rectangle? I always get nowhere with the semicircle.
     
  2. jcsd
  3. Nov 29, 2009 #2
    the limits are from c - infinity*i to c + infinity*i where c is a constant bigger than zero.
     
  4. Nov 29, 2009 #3

    HallsofIvy

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    It would help a lot if you would tell us what the problem really is!
     
  5. Nov 30, 2009 #4
    prove:

    x^s/s integrated from c-infinity*i to c+infinity*i, where c is any constant bigger than zero, is equal to 1 if x is bigger than 1. The outline of the proof is that we use a semicircle contour enclosing the origin, whose line segment parallel to the imaginary axis is intersecting the real axis at x=c and the semicircle part goes to the left obviously, if it encloses the origin, since c is positive. We then proceed to show that the total contour integral equals 1, and the semi circle part equals zero, giving the desired result. The part i am having difficulty in is proving that the semicircle part goes to zero. any help would be greatly appreciated.
     
  6. Nov 30, 2009 #5
    Should that be "if s is bigger than 1"?
     
  7. Dec 1, 2009 #6
    no, s is the variable we are integrating. The integrand, including the dummy variable, is x^s/sds.
     
  8. Dec 2, 2009 #7
    So, this is an inverse Laplace transform problem. We know that the Laplace transform of the constant function equal to 1 is 1/s, so transforming it back should yield 1. You want to verify this via direct computation.

    THe integral is than as you said it is, but with a 1/(2 pi i) factor in front of it. If you close the contour with the semi-circle, then you can directly apply the residue theorem. There is a pole at s = 0, the residue there is 1.
     
  9. Dec 2, 2009 #8
    Yes, but can you show me how the semicircle goes to zero?
     
  10. Dec 2, 2009 #9
    You can write x^s as exp(s t) with t = Log(x) > 0. The proof that the semi-circle goes to zero is practically the same as in the case of the integral of sin(x)/x.
     
  11. Dec 2, 2009 #10
    Oh, it comes from jordans lemma, thank you! I guess i was caught up on this problem because the book problems in analytic number theory by murty doesnt thoroughly explain it, at least to my satisfaction. I think its mistaken, though. If you have it, could you please take a look at it and give me your insight on why I am incorrect?
     
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