 Solution

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

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

In the integral defining the Beta function, put

t = (1 + x) /2

and then

u = x^{2}

B (z, z) = 2^{1 - 2 z} \int_{- 1}^{1} (1 - x^{2})^{z - 1} d x = 2^{2 - 2 z} \int_{0}^{1} (1 - x^{2})^{z - 1} d x = 2^{1 - 2 z} \int_{0}^{1} u^{- 1/2} (1 - u)^{z - 1} d u = 2^{1 - 2 z} B (z, \frac{1}{2}) .

This proves the identity for

Re z > 0

. Both sides extend by analytic continuation to the same meromorphic function, equivalently by the beta--gamma identity and the Gamma duplication formula. The identity holds meromorphically for every $z \in C$ ; both sides are finite for $z \in / {0, - 1, - 2, \dots}$ and have matching poles at the excluded points.

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