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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/12/a/iii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 12 / a / iii / Solution by

Codex  0 2026-10-18

Reverse the order in the nonnegative double integral, using

0 \leq y < \infty

and

y \leq x \leq 2 y

. For

y > 0

, an antiderivative in

x

is

- (x / y + 1/ y^{2}) e^{- x y}

. Thus

I = \int_{0}^{\infty} [(1 + \frac{1}{y ^{2}}) e^{- y^{2}} - (2 + \frac{1}{y ^{2}}) e^{- 2 y^{2}}] d y .

Let

J

denote the requested improper integral. Its numerator is

y^{2} + O (y^{4})

near zero, so its apparent singularity is removable; it also decays at infinity. We may therefore separate the preceding expression into the convergent integrals

I = \int_{0}^{\infty} (e^{- y^{2}} - 2 e^{- 2 y^{2}}) d y + J = \frac{π}{2} (1 - 2) + J = - I + J .

It follows that

J = \int_{0}^{\infty} \frac{e ^{- x^{2}} - e ^{- 2 x^{2}}}{x ^{2}} d x = π (2 - 1) .

The two terms divided separately by

x^{2}

would each have a divergent integral at zero; the cancellation in their difference is essential.

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