Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 327 / 1 / b / iv

Table of contents
- Solution iv

Solution

On either open half-line, part (i) says

u = - v^{'}

. Differentiating the formula in part (iii) makes all elementary terms cancel except the logarithm and the bracket. Therefore

u (λ) = - lo g ∣ λ ∣ + f_{+} (λ), λ > 0,

where

f_{+} (λ) = \int_{λ}^{1} \frac{e ^{- i x} - 1}{x} d x + \int_{1}^{\infty} \frac{e ^{- i x}}{x} d x,

and

u (λ) = - lo g ∣ λ ∣ + f_{-} (λ), λ < 0,

where

f_{-} (λ) = \int_{∣ λ ∣}^{1} \frac{e ^{i x} - 1}{x} d x + \int_{1}^{\infty} \frac{e ^{i x}}{x} d x .

Both functions are smooth on their respective half-lines, proving the required assertion.

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