Past exam of the options I course of the University of Oxford / 2004 / D / D2

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- Solution D2

Solution

The indicated measurable function is nonnegative. Tonelli's theorem applies without first assuming integrability and gives

\int_{R^{2}} h = \int_{a}^{b} (\int_{1}^{\infty} x^{- y} d x) d y = \int_{a}^{b} \frac{d y}{y - 1} = lo g \frac{b - 1}{a - 1} < \infty.

Thus

h

is integrable. Fubini's theorem now allows the reverse order. For

x > 1

,

\int_{a}^{b} x^{- y} d y = \frac{x ^{- a} - x ^{- b}}{l o g x} .

Consequently

\int_{1}^{\infty} \frac{x ^{- a} - x ^{- b}}{lo g x} d x = lo g \frac{b - 1}{a - 1} .

At

x = 1

the quotient has the removable limit

b - a

; that endpoint does not create an additional divergence.

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