Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2007/6/ii/solution

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 , with
The 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 to
The residue theorem therefore gives
Taking real parts yields

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