Past exam of the integration course of the University of Oxford / 2021 / 1 / a

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- Solution a

Solution

Extend the integrand by zero to all of

(0, \infty)

:

h_{n} (x) = 1_{(0, 2 nπ)} (x) \frac{( n + x ) s i n ( x / n )}{x ( 1 + x ^{2} )} .

For each fixed

x > 0

, eventually the indicator function is one and

sin (x / n) / (x / n) \to 1

. Hence

h_{n} (x) \to (1 + x^{2})^{- 1}

pointwise. On its support,

x / n \leq 2 π

, so

∣ sin t ∣ \leq t

for

t \geq 0

gives

∣ h_{n} (x) ∣ \leq \frac{1 + x / n}{1 + x ^{2}} \leq \frac{1 + 2 π}{1 + x ^{2}} .

The final bound is Lebesgue integrable on

(0, \infty)

and independent of

n

. The dominated convergence theorem therefore gives

n \to \infty lim \int_{0}^{2 nπ} \frac{( n + x ) sin ( x / n )}{x ( 1 + x ^{2} )} d x = \int_{0}^{\infty} \frac{d x}{1 + x ^{2}} = \frac{π}{2} .

The last integral follows from the antiderivative

arctan x

.

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