 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2010/d/d4/2/solution

A direct derivative bound by

1/ x

would not be integrable, so the earlier theorem cannot be applied that way. Instead, for

t > 0

make the legitimate substitution

s = t x

K (t) = t \int_{t}^{\infty} \frac{c o s s}{s ^{2}} d s .

The tail integral has continuous integrand away from zero and converges absolutely. The fundamental theorem of calculus therefore gives derivative

- cos t / t^{2}

with respect to its lower endpoint. The product rule yields

K^{'} (t) = \int_{t}^{\infty} \frac{cos s}{s ^{2}} d s - \frac{cos t}{t} = \frac{K ( t ) - cos t}{t} .

This establishes differentiability carefully without an invalid domination argument.

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