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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-79/1/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 79 / 1 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

The complete integral is

Γ (1/2) = π

. Therefore apply asymptotic expansion by repeated integration by parts to the tail

I (x) = \int_{x}^{\infty} t^{- 1/2} e^{- t} d t

:

I (x) = e^{- x} x^{- 1/2} - \frac{1}{2} \int_{x}^{\infty} t^{- 3/2} e^{- t} d t = e^{- x} (x^{- 1/2} - \frac{1}{2} x^{- 3/2} + \frac{3}{4} x^{- 5/2}) - \frac{15}{8} \int_{x}^{\infty} t^{- 7/2} e^{- t} d t .

The last integral is bounded by

x^{- 7/2} e^{- x}

. Thus

f (x) = π - e^{- x} x^{- 1/2} + \frac{1}{2} e^{- x} x^{- 3/2} - \frac{3}{4} e^{- x} x^{- 5/2} + O (e^{- x} x^{- 7/2}) .

In particular the first three additive terms are

π

,

- e^{- x} x^{- 1/2}

and

\frac{1}{2} e^{- x} x^{- 3/2}

; the formula also supplies three successive terms of the exponentially small tail if the constant is counted separately. A power expansion alone would retain only

π

and miss this information.

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