Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 336 / 2 / a

Solution

Let

D (x) = x (x + ε) + ε^{3} e^{- x^{2}}, D_{0} (x) = x (x + ε) + ε^{3} .

The inequality

∣1 - e^{- x^{2}} ∣ \leq x^{2}

and a split into

0 < x < ε^{2}

ε^{2} < x < ε

, and

ε < x < 1

show that

\int_{0}^{1} \frac{1}{D} - \frac{1}{D _{0}} d x \leq C \int_{0}^{1} \frac{ε ^{3} x ^{2}}{D _{0} ( x ) ^{2}} d x = O (ε^{2}) .

Thus replacing the exponential by one does not affect any term through

O (1)

Factor

D_{0} (x) = (x + a) (x + b),

where

a = \frac{ε}{2} (1 - 1 - 4 ε), b = \frac{ε}{2} (1 + 1 - 4 ε) .

\int_{0}^{1} \frac{d x}{D _{0} ( x )} = \frac{1}{ε 1 - 4 ε} [lo g \frac{b}{a} + lo g \frac{1 + a}{1 + b}] .

The required asymptotic expansion follows from

\frac{1}{1 - 4 ε} = 1 + 2 ε + O (ε^{2}),

lo g \frac{b}{a} = - lo g ε - 2 ε + O (ε^{2}), lo g \frac{1 + a}{1 + b} = - ε + O (ε^{2}) .

Therefore

I (ε) = \frac{lo g ( 1/ ε )}{ε} + 2 lo g (1/ ε) - 3 + o (1) .

The two small roots reveal the same nested scales

x = O (ε^{2})

and

x = O (ε)

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