 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/30a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Let

I (x) = \int_{x}^{\infty} e^{- t^{2}} d t .

Since

d (e^{- t^{2}}) = - 2 t e^{- t^{2}} d t

I (x) = \frac{e ^{- x^{2}}}{2 x} - \frac{1}{2} \int_{x}^{\infty} t^{- 2} e^{- t^{2}} d t = \frac{e ^{- x^{2}}}{2 x} - \frac{e ^{- x^{2}}}{4 x ^{3}} + \frac{3}{4} \int_{x}^{\infty} t^{- 4} e^{- t^{2}} d t = \frac{e ^{- x^{2}}}{2 x} (1 - \frac{1}{2 x ^{2}} + \frac{3}{4 x ^{4}} + O (x^{- 6})) .

The final remainder estimate follows by one further integration by parts, or by bounding the remaining positive integral by its leading boundary term. Therefore the requested first three terms are

erfc (x) \sim \frac{e ^{- x^{2}}}{2 2 π x} (1 - \frac{1}{2 x ^{2}} + \frac{3}{4 x ^{4}}) .

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