 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/16/a/solution

For

p > 0

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

The endpoint terms vanish because

x^{p} \to 0

at zero and the exponential dominates the power at infinity. Also

Γ (1) = \int_{0}^{\infty} e^{- x} d x = 1

. Iterating for nonnegative integers

n

therefore yields

Γ (n + 1) = n! .

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