 Solution

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

A naive absolute bound on the original derivative would be

1/ x

, which is not integrable on

[1, \infty)

. Instead make the change of variable

s = t x

, valid for

t > 0

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

The integral is absolutely convergent for every positive lower endpoint. Its integrand is continuous there, so the fundamental theorem of calculus gives

\frac{d}{d t} \int_{t}^{\infty} \frac{c o s s}{s ^{2}} d s = - \frac{c o s t}{t ^{2}} .

The product rule now yields

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

This proves differentiability and the equation without an unjustified differentiation of a conditionally convergent derivative integral.

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