 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2007/d/d4/ii/solution

At each

x > 0

, eventually

x < n

, and the stipulated power limit gives

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

. At nonpositive

x

the functions vanish. On their positive support,

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

The supplied integrable exponential therefore dominates the sequence. Dominated convergence gives

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

The final improper integral is absolutely convergent, so agrees with its Lebesgue integral.

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