What Happens to the Integral of Cosine as k Approaches Infinity?

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    Cosine Infinity
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Discussion Overview

The discussion revolves around the evaluation of the improper integral of cosine, specifically the integral 2∫cos(kx)dk from zero to infinity. Participants explore whether this integral converges and how to describe its behavior as k approaches infinity.

Discussion Character

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

Main Points Raised

  • Some participants note that the integral 2∫cos(kx)dk from zero to infinity does not converge.
  • Others emphasize the need to use limits when evaluating improper integrals, particularly when the upper limit is infinity.
  • A participant suggests that the limit as k approaches infinity would involve sin(kx), questioning how to evaluate it.
  • Another participant points out that the limit does not exist due to the oscillatory nature of sin(kx).

Areas of Agreement / Disagreement

Participants generally agree that the integral does not converge and that the oscillatory behavior of sin(kx) complicates the evaluation of the limit. However, there is no consensus on a definitive method for describing the integral's behavior.

Contextual Notes

The discussion highlights the challenges associated with improper integrals and the necessity of using limits, but lacks specific resolutions or established methods for evaluation.

maverick_76
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okay so I have this integral:

2∫cos(kx)dk

The bounds are from zero to infinity, this doesn't converge but is there any other way to describe this?
 
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maverick_76 said:
okay so I have this integral:

2∫cos(kx)dk

The bounds are from zero to infinity, this doesn't converge but is there any other way to describe this?
Like this? ##2\int_0^{\infty}\cos(kx) dk##
Because the upper limit of integration is ∞, this is an improper integral. You can't just "plug in" ∞ when you evaluate the antiderivative -- you need to use limits to evaluate it.
 
okay so the limit would be lim_k->∞ sin(kx)
xHow would I go about evaluating it? Is there a substitution trick?
 
maverick_76 said:
okay so the limit would be lim_k->∞ sin(kx)
xHow would I go about evaluating it? Is there a substitution trick?
The limit doesn't exist because sin(kx) oscillates endlessly.
 
okay gotcha. Thanks!
 

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