Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 20 / c

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- Solution c

Solution

The parameter derivative of the integrand is

cos (a lo g x)

, whose absolute value is bounded by one on

0 < x < 1

. This integrable bound justifies differentiation under the integral sign, giving

I^{'} (a) = \int_{0}^{1} cos (a lo g x) d x .

The original integral also exists:

∣ sin (a lo g x) / lo g x ∣ \leq ∣ a ∣

, and its apparent singularity at

x = 1

is removable. Set

x = e^{- t}

to evaluate the derivative:

I^{'} (a) = \int_{0}^{\infty} e^{- t} cos (a t) d t = Re \int_{0}^{\infty} e^{- (1 - ia) t} d t = Re \frac{1}{1 - ia} .

Consequently

I^{'} (a) = \frac{1}{1 + a ^{2}} .

Since

I (0) = 0

, integration in the parameter fixes the constant and gives the logarithmic sine integral identity

I (a) = arctan a, I (1) = \frac{π}{4} .

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