 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2007/d/4/ii/solution

For fixed

x > 0

f_{n} (x) \to e^{- x} sin x

; for

x \leq 0

, the limit is zero. On

0 < x < n

, the supplied inequality gives

(1 - x / n)^{n} \leq e^{- x}

, so

∣ f_{n} (x) ∣ \leq χ_{(0, \infty)} (x) e^{- x} .

This is the given integrable dominating function. By the dominated convergence theorem,

lim_{n} \int f_{n} (x) d x = \int_{0}^{\infty} e^{- x} sin x d x = Im \frac{1}{1 - i} = \frac{1}{2} .

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