Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / D / 4 / b / i

Table of contents
- Solution i

Solution

For fixed

n > 0

, Tonelli gives

\int_{0}^{n} \int_{0}^{\infty} ∣ sin x ∣ e^{- t x} d t d x = \int_{0}^{n} \frac{∣ s i n x ∣}{x} d x < \infty,

because the quotient is bounded near zero. Fubini therefore permits reversing the order for this finite

x

interval. Direct integration in

x

yields

\int_{0}^{n} e^{- t x} sin x d x = \frac{1 - e ^{- n t} ( t s i n n + c o s n )}{1 + t ^{2}} .

Thus the desired integral equals

π /2

minus the integral of the exponential remainder. Its absolute value is at most

\int_{0}^{\infty} e^{- n t} (t + 1) d t = \frac{1}{n ^{2}} + \frac{1}{n} ⟶ 0.

Consequently

n \to \infty lim \int_{0}^{n} sin x / x d x = π /2

. The truncation is essential: Fubini is not being applied to an absolutely integrable function over the whole infinite rectangle.

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