Is the Dirichlet integral a shortcut for solving this difficult integral?

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Discussion Overview

The discussion revolves around the evaluation of the integral ##\displaystyle \int_{- \infty}^{\infty} \frac{\cos x}{x^2+1} dx## using differentiation under the integral sign and its potential connection to the Dirichlet integral. Participants explore various approaches to manipulate the integral, including variable substitution and the use of properties of even functions.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant introduces the integral and the function ##I(t) = \int_{- \infty}^{\infty} \frac{\cos tx}{x^2+1} dx##, questioning the connection to the Dirichlet integral due to the presence of ##t## in the argument of ##\sin##.
  • Another participant suggests changing variables to simplify the integral, proposing ##u=tx## as a potential method.
  • A follow-up comment expresses uncertainty about the sign of ##t## and its effect on the limits of integration, indicating a concern about whether the upper bound approaches ##+\infty## or ##-\infty##.
  • One participant argues that the final answer should not depend on the sign of ##t## because ##I(t)=I(-t)##, noting the even nature of the functions involved.
  • A question is raised regarding the use of calculus of residues as an alternative method to evaluate the original integral.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to evaluate the integral, with some favoring variable substitution and others considering the properties of even functions. The discussion remains unresolved regarding the optimal method and the implications of the sign of ##t##.

Contextual Notes

Participants highlight the dependence on the sign of ##t## and the implications for the limits of integration, as well as the potential link to the Dirichlet integral, which remains to be fully clarified.

Mr Davis 97
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I have the integral ##\displaystyle \int_{- \infty}^{\infty} \frac{\cos x}{x^2+1} dx##. We are going to use differentiation under the integral sign, so we let ##\displaystyle I(t) = \int_{- \infty}^{\infty} \frac{\cos tx}{x^2+1} dx##, and then, after manipulation, ##\displaystyle I'(t) = \int_{- \infty}^{\infty} \frac{\sin tx}{x(x^2+1)} dx - \int_{- \infty}^{\infty} \frac{\sin tx}{x} dx##. My question lies in the rightmost integral. In a solution I've seen, the rightmost integral is linked to the Dirichlet integral: https://en.wikipedia.org/wiki/Dirichlet_integral. And so ##\pi## is simply substituted for this expression. What I don't understand is how can it be linked to this known integral when there is that ##t## in the argument of ##\sin##?
 
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Can you not change variables? Let ##u=tx##, etc. etc.
 
kuruman said:
Can you not change variables? Let ##u=tx##, etc. etc.
Okay, I see how that could work... But I don't know the sign of t, right? As such I can't tell whether the upper bound on the integral goes to ##+ \infty## or ##- \infty##
 
The final answer should not depend on the sign of ##t## because ##I(t)=I(-t)##. You have an even function in ##x## and ##t##. That helps you to link to the Dirchlet integral by putting a ##2## up front and taking the limits from zero to infinity.
 
Is there a reason for why you don't deal with the original integral via calculus of residues?
 

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