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

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Discussion Overview

The discussion revolves around proving that the integral of the function x^s/s, integrated from c - infinity*i to c + infinity*i (where c is a constant greater than zero), equals 1 when x is greater than 1. The focus is on using a semicircle contour for the proof, particularly addressing challenges related to the semicircle part of the contour integral.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant requests a proof using a semicircle contour instead of a rectangle, expressing difficulty with the semicircle approach.
  • Another participant clarifies the limits of integration and the conditions for the proof.
  • A participant questions the clarity of the problem statement, suggesting that more context is needed.
  • One participant outlines the proof structure, emphasizing the semicircle contour and the need to show that the semicircle part of the integral approaches zero.
  • There is a suggestion that the condition should involve s being greater than 1, which is clarified to be incorrect as s is the variable of integration.
  • A participant identifies the problem as related to the inverse Laplace transform and discusses the application of the residue theorem, noting a pole at s = 0.
  • Another participant seeks clarification on how the semicircle part approaches zero.
  • One participant proposes expressing x^s as exp(s t) with t = Log(x) and relates the proof to a known result involving the integral of sin(x)/x.
  • A participant expresses gratitude for identifying Jordan's lemma as a key to understanding the semicircle's behavior and mentions dissatisfaction with the explanations in a referenced book.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to the proof, with some uncertainty about the semicircle's contribution to the integral. There is no consensus on the best method to demonstrate that the semicircle part goes to zero, and some participants propose different perspectives on the problem.

Contextual Notes

The discussion includes assumptions about the behavior of the semicircle contour and the applicability of Jordan's lemma, which may not be fully resolved. The participants also reference specific mathematical techniques without reaching a definitive conclusion on the proof's validity.

jack5322
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can someone prove this for me using a semicircle contour as opposed to a rectangle? I always get nowhere with the semicircle.
 
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the limits are from c - infinity*i to c + infinity*i where c is a constant bigger than zero.
 
It would help a lot if you would tell us what the problem really is!
 
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.
 
Should that be "if s is bigger than 1"?
 
no, s is the variable we are integrating. The integrand, including the dummy variable, is x^s/sds.
 
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.
 
Yes, but can you show me how the semicircle goes to zero?
 
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.
 
  • #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 doesn't 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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