Complex Integrals: Evaluating \int_{-1}^1 \frac{(1-x^2)^{1/2}}{x^2+1}dx

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Homework Help Overview

The discussion revolves around evaluating the integral \(\int_{-1}^1 \frac{(1-x^2)^{1/2}}{x^2+1}dx\) using complex analysis techniques, specifically contour integration. Participants are exploring the implications of using the complex function \(\frac{(z^2-1)^{1/2}}{z^2+1}\) in this context.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are questioning the relevance of the factor of \(i\) in the complex function and its impact on the final result. There is speculation about whether the focus should be on the real part of the integral and how the presence of imaginary components might affect the evaluation.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts on the relationship between the complex integral and the original real integral. Some have suggested that the imaginary parts may not matter in the final evaluation, while others express uncertainty about how these factors interact.

Contextual Notes

There is a concern regarding the differences between the complex and real parts of the integral, particularly in relation to the presence of \(y\) in the complex variable \(z = x + iy\). Participants are also considering the implications of residue cancellation in the context of the integral.

Tangent87
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In our Complex Methods lecture today, our lecturer went through the example of evaluating the integral [tex]\int_{-1}^1 \frac{(1-x^2)^{1/2}}{x^2+1}dx[/tex] and then proceeded to do the whole contour calculation using the complex function [tex]\frac{(z^2-1)^{1/2}}{z^2+1}[/tex]. I'm concerned that the answer will be a factor of i out because [tex](1-z^2)^{1/2}=i(z^2-1)^{1/2}[/tex] but he said in the lecture it didn't matter, can anyone explain why?
 
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Maybe because in the end you are only interested in real solution (part)?

Just guessing here...
 
dingo_d said:
Maybe because in the end you are only interested in real solution (part)?

Just guessing here...

You're probably on the right lines but the real part of [tex](z^2-1)^{1/2}[/tex] is def not [tex](1-x^2)^{1/2}[/tex] because at the very least there'll be loads of y's involved as well since z=x+iy?
 
Oh, I was thinking that at the end of the integration you get something real and sth imaginary and then you take the real part...

Don't take that as a concrete answer, I too don't see how that wouldn't matter at first...

Because it really is not the same integral (by that factor).

Altho the residues do seem to cancel each other, so maybe there's the catch :\
 

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