 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ib/paper-2/14f/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Close the real-line contour by a positively oriented upper semicircle of radius

R

, beyond every pole. The rational function is

O (R^{- 2})

uniformly on that arc, and

∣ e^{i z} ∣ = e^{- ℑ z} \leq 1

. Its arc integral is therefore

O (R^{- 1})

and vanishes. The real integral is absolutely convergent, while no pole lies on it. The residue theorem gives

\int_{- \infty}^{\infty} F (x) e^{i x} d x = 2 πi ℑ z_{j} > 0 \sum Res_{z = z_{j}} (F (z) e^{i z}) .

Only actual poles contribute; any removable zero of the denominator has zero residue. The same contour proof covers poles of any order.

For the particular denominator, the upper poles are

a = 1 + i

and

b = - 1 + i

. They are simple, with residues

e^{ia} / (4 a^{3})

and

e^{ib} / (4 b^{3})

. Since

\frac{1}{4 a ^{3}} = \frac{- 1 - i}{16}, \frac{1}{4 b ^{3}} = \frac{1 - i}{16},

their sum is

\frac{e ^{- 1}}{16} [(- 1 - i) e^{i} + (1 - i) e^{- i}] = - \frac{i e ^{- 1}}{8} (cos 1 + sin 1) .

2 πi

gives the real answer. The sine part of the real-line integral is zero by oddness, so

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

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