Complex Integral Trigonometric Substitution

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Discussion Overview

The discussion centers on the use of complex analysis and the residue theorem to evaluate definite integrals involving trigonometric functions, specifically \(\frac{\sin(z)}{z}\) and \(\frac{\cos(z)}{e^z + e^{-z}}\). Participants explore the justification for replacing trigonometric functions with their exponential forms in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the validity of replacing \(\sin(z)\) and \(\cos(z)\) with \(e^{iz}\) in the context of complex integrals and the residue theorem.
  • Another participant provides the identities for \(\sin(z)\) and \(\cos(z)\) in terms of exponential functions, indicating familiarity with these transformations.
  • A different participant suggests that the replacement may be related to the geometry of the problem and the singularity at \(z=0\), implying a deeper reasoning behind the transformation.
  • Another reply emphasizes the importance of analyzing the paths of the integrals and how the closed path integral can inform the evaluation of integrals along the real line, suggesting a method to separate real and imaginary parts for the trigonometric functions.

Areas of Agreement / Disagreement

Participants express differing views on the justification for the transformation of trigonometric functions into exponential forms. There is no consensus on the reasoning behind this approach, and the discussion remains unresolved.

Contextual Notes

Participants acknowledge the singularity at \(z=0\) and the geometry of the integral paths, but the implications of these factors on the transformation remain unclear. The discussion does not resolve the mathematical steps involved in the integral evaluation.

KleZMeR
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I'm solving two different definite integrals of functions

[itex]\frac{sin(z)}{z}[/itex] and [itex]\frac{cos(z)}{e^z+e^{-z}}[/itex]

with complex analysis and the residue theorem, and in the solutions they replace both

[itex]sin(z)[/itex] and [itex]cos(z)[/itex] with [itex]e^{iz}[/itex]

why is this possible?
 
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sin(z) = (e^{iz}-e^{-iz})/2i, cos(z) = (e^{iz}+e^{-iz})/2
 
Thanks Mathman, but I know of these identities. I was not asking about replacing trig with polar form. Because of the nature of the Residue Theorem and the geometry of the problem, for some reason I think they have decomposed the regions because of the singularity at z=0, and as a result the integral changes to a single term as stated above.
 
Take a close look at the paths of the integrals. On the real line, eir = cos(r) + i*sin(r). So if you know the total integral along the closed path and the values of the integrals off of the real line, you can determine the integral on the integral path part on the real line. Then you can pull off the real and imaginary parts to get the integrals of the cos and sin, respectively.
 
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