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

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SUMMARY

The forum discussion centers on evaluating the complex integral \(\int_{-1}^1 \frac{(1-x^2)^{1/2}}{x^2+1}dx\) using contour integration with the function \(\frac{(z^2-1)^{1/2}}{z^2+1}\). Participants express concerns about the potential introduction of an imaginary factor due to the expression \((1-z^2)^{1/2} = i(z^2-1)^{1/2}\). The lecturer clarified that the focus is on obtaining the real part of the integral, which remains unaffected by the imaginary component introduced during the calculation. The discussion highlights the importance of understanding how residues in complex analysis can cancel each other, impacting the final result.

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Tangent87
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In our Complex Methods lecture today, our lecturer went through the example of evaluating the integral \int_{-1}^1 \frac{(1-x^2)^{1/2}}{x^2+1}dx and then proceeded to do the whole contour calculation using the complex function \frac{(z^2-1)^{1/2}}{z^2+1}. I'm concerned that the answer will be a factor of i out because (1-z^2)^{1/2}=i(z^2-1)^{1/2} 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 (z^2-1)^{1/2} is def not (1-x^2)^{1/2} 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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