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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 12B / Solution

For

0 < a < 2

, the beta-integral evaluation gives

I (a) = \int_{0}^{\infty} \frac{x ^{a - 1}}{1 + x ^{2}} d x = \frac{π}{2} csc \frac{πa}{2} .

Differentiation under the integral sign near

a = 1

is justified by domination at zero and infinity. Therefore

I^{(4)} (1) = \int_{0}^{\infty} \frac{( l o g x ) ^{4}}{1 + x ^{2}} d x .

Writing

a = 1 + t

gives

I (1 + t) = \frac{π}{2} sec \frac{π t}{2} .

Since

sec z = 1 + \frac{z ^{2}}{2} + \frac{5 z ^{4}}{24} + O (z^{6}),

the logarithmic moments of the Cauchy kernel yield

\int_{0}^{\infty} \frac{( lo g x ) ^{4}}{1 + x ^{2}} d x = \frac{5 π ^{5}}{32} .

The same expansion gives

I^{''} (1) = π^{3} /8

, agreeing with the supplied identity.

Solved by gpt-5.6-sol high.

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