Evaluating integrals using the residue theorem

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SUMMARY

This discussion focuses on evaluating integrals using the residue theorem, specifically the integral \(\int_{-\infty}^{\infty} \frac{dx}{\sqrt{x^2 + 2}(x^2+1)}\), which evaluates to \(\frac{\pi}{2}\). The user, Sven, encounters confusion regarding the factor of 1/2 in the result, despite correctly identifying the residue at the pole +i as \(-i/2\). The discussion also touches on the potential impact of branch cuts and the Riemann sphere on the evaluation of integrals, with suggestions to utilize the Weierstraß or tangent half-angle substitution for simplification.

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sschmidt
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Hi,

I have trouble evaluating simple integrals like

[tex] \int_{-\infty}^{\infty} \frac{dx}{\sqrt{x^2 + 2}(x^2+1)} = \frac{\pi}{2}[/tex]
I'd like to calculate the integral closing the integration loop in the upper half-plane enclosing the pole at +i. The residue is - i / 2 and hence 2 \pi i ( - i \ 2 ) is \pi, but the integral should evaluate to \pi / 2. I just don't seem to see what is happening here. The other pole at - i shouldn't interfere and the residue at +i is not evaluated twice, so where does the missing factor 1/2 come from? Is this related to the Riemann sphere and its branch cuts? Does the contour with semicircle cross branch cuts and change the leaf?
What are the branch cuts here anyway? The same problem occurs with

[tex] \int_{-\infty}^{\infty} \frac{dx}{\sqrt{x^2 + 4}(x^2+1)} = \frac{2}{3} \pi i (-\frac{i}{\sqrt{3}})[/tex]

Thank you very much and kind regards, Sven
 
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