Integral using parseval (supposed to be simple )

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SUMMARY

The discussion centers on evaluating the integral of the function \( \frac{1}{a^2 + w^2} \) from \(-a\) to \(a\). The correct evaluation of this integral is \( \int \frac{1}{a^2 + w^2} dw = \frac{1}{a} \tan^{-1} \frac{w}{a} + C \). The user expresses confusion regarding the steps leading to this conclusion and seeks clarification on the application of Parseval's theorem in this context.

PREREQUISITES
  • Understanding of integral calculus, specifically definite integrals.
  • Familiarity with the arctangent function and its properties.
  • Knowledge of Parseval's theorem and its relevance in Fourier analysis.
  • Basic algebraic manipulation skills for simplifying integrals.
NEXT STEPS
  • Study the derivation of the integral \( \int \frac{1}{a^2 + w^2} dw \).
  • Learn about Parseval's theorem and its applications in signal processing.
  • Explore techniques for evaluating definite integrals involving trigonometric functions.
  • Review the properties of the arctangent function and its integral representation.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and Fourier analysis, will benefit from this discussion. It is also useful for anyone looking to deepen their understanding of integral evaluation techniques.

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integral using parseval (supposed to be simple :( )

Hi everyone. I am stuck in one of the assignment problems, and I think it's supposed to be simple, but it's just that I don't see how it worked out.

I have to take integral(1/a^2+w^2)dw, from -a to a.
I know the answer but I have no single clue how it came out.
It would be great if someone could explain with steps :(
Thank you and have a great day.
 
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Do you mean this?
\int \frac{1}{a^2+w^2}dw = \frac{1}{a}\tan^{-1}\frac{w}{a}+c

or this?

\int \left(\frac{1}{a^2}+{w^2}\right)dw = \ldots

Why Parseval?
 

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