Qualifying Examination
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چکیده
Solution. We can consider the integration from −∞ to ∞ instead. For R > 1, consider the contour that consists of the segment from −R to R and the arc {Reiθ | θ ∈ [0, π]}. Since z2+1 z4+1 decays as |z|−2 on the complex plane as |z| → ∞, this contour integral converges to twice of the original integral when R→∞. The contour encloses two simple poles eπi/4 and e3πi/4 of the function. At eπi/4 the residue of the function is (e πi/4)2+1 d dz (z4+1)| z=eπi/4 = 1+i 4e3πi/4 = −i 2 √ 2 . Similarly
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