 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2010/analysis/6/iii/solution

w = i

, which satisfies

ℜ w = 0

, and take real parts of the absolutely convergent integral. Since

x^{i} = cos (lo g x) + i sin (lo g x)

and

cos (iπ /2) = cosh (π /2)

\int_{0}^{\infty} \frac{cos ( lo g x )}{1 + x ^{2}} d x = \frac{π}{2 cosh ( π /2 )} .

Absolute integrability justifies moving the real-part operation through the integral.

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