Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 30A / b

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- Solution b

Solution

Let

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

Since

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

, repeated integration by parts gives

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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