 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/17/b/solution

For

n > 1

u = x^{n - 1}

and

d v = e^{- x} d x

yields

I_{n} = [- x^{n - 1} e^{- x}]_{0}^{\infty} + (n - 1) \int_{0}^{\infty} x^{n - 2} e^{- x} d x = (n - 1) I_{n - 1} .

The boundary term vanishes at zero because

n - 1 > 0

, and at infinity because exponential decay dominates every fixed power. The initial improper integral is

I_{1} = \int_{0}^{\infty} e^{- x} d x = 1

. Iterating the recurrence relation therefore gives

I_{n} = (n - 1) (n - 2) \dots 1 I_{1} = (n - 1)!, n \geq 1.

This is the positive-integer case of the Gamma function identity

Γ (n) = (n - 1)!

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