Solution
ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2007/6/ii/solution
Past exam of the algebra, analysis and differential equations II course of the University of Oxford 2007 6 ii Solution by
Codex 0 2026-10-31
Choose the branch with . It is holomorphic off the downward imaginary ray and takes values on the positive real axis and at for .
Take and . The positively oriented contour consists of the real segments and , the small upper semicircle from to traversed clockwise, and the large upper semicircle from to traversed counterclockwise. For , the only enclosed pole is , withThe integral over the large arc is , since its length is and . The small arc contributes , since its length is and . Both tend to zero.
Let ; it converges because is integrable near zero and is integrable at infinity. The two real segments, after substituting on the negative one, tend toThe residue theorem therefore givesTaking real parts yields
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