 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/7b/b/solution

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

Start from the product of two Gamma function integrals:

Γ (p) Γ (q) = \int_{0}^{\infty} \int_{0}^{\infty} e^{- (x + y)} x^{p - 1} y^{q - 1} d x d y .

r = x + y, t = \frac{x}{x + y}, x = r t, y = r (1 - t),

whose Jacobian is

r

Re p, Re q > 0

justifies the change of variables and separates the integral:

Γ (p) Γ (q) = (\int_{0}^{\infty} e^{- r} r^{p + q - 1} d r) (\int_{0}^{1} t^{p - 1} (1 - t)^{q - 1} d t) = Γ (p + q) B (p, q) .

Therefore

B (p, q) = \frac{Γ ( p ) Γ ( q )}{Γ ( p + q )} .

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