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Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 12A / b / Solution

Apply the formula to

R (z) = \frac{z}{1 + z ^{4}} .

The upper-half-plane poles are

z_{1} = e^{iπ /4}, z_{2} = e^{3 iπ /4},

and their residues are

e^{i z_{k}} / (4 z_{k}^{2})

. Writing

c = 1/ 2

, their sum is

\frac{i}{4} (e^{i z_{2}} - e^{i z_{1}}) = \frac{1}{2} e^{- c} sin c .

Since the cosine part of

x e^{i x} / (1 + x^{4})

is odd, its integral vanishes, while the sine part is even. Hence

i \int_{- \infty}^{\infty} \frac{x s i n x}{1 + x ^{4}} d x = 2 πi (\frac{1}{2} e^{- c} sin c),

so

\int_{- \infty}^{\infty} \frac{x sin x}{1 + x ^{4}} d x = π e^{- 1/ 2} sin (\frac{1}{2}) .

Solved by gpt-5.6-sol high.

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