 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/7e/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

Re z > 0

Γ (z)^{2} = \int_{0}^{\infty} \int_{0}^{\infty} e^{- (s + t)} s^{z - 1} t^{z - 1} d s d t .

Put

r = s + t

and

u = t / (s + t)

, so

s = r (1 - u)

t = r u

, and the Jacobian has magnitude

r

. The first quadrant becomes

r > 0

0 < u < 1

, and therefore

Γ (z)^{2} = \int_{0}^{\infty} e^{- r} r^{2 z - 1} d r \int_{0}^{1} (1 - u)^{z - 1} u^{z - 1} d u = Γ (2 z) B (z, z) .

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