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 Incoming links: Beta--gamma identity

Beta-function recurrence 2026-09-28

The Beta function obeys

B (x + 1, y) = \frac{x}{x + y} B (x, y), B (x, y + 1) = \frac{y}{x + y} B (x, y) .

These identities follow immediately from the beta--gamma identity and the Gamma function recurrence.

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Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 14E / b / Solution Created 2026-09-24 Updated 2026-09-24

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Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 8E / Solution Created 2026-09-24 Updated 2026-09-24

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Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 7B / a / Solution 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}

. Symmetry gives

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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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 7A / a / Solution Created 2026-09-24 Updated 2026-09-29

With

u = x^{4}

, the change of variables formula gives

\int_{0}^{1} \frac{d x}{1 - x ^{4}} = \frac{1}{4} \int_{0}^{1} u^{- 3/4} (1 - u)^{- 1/2} d u = \frac{1}{4} B (\frac{1}{4}, \frac{1}{2}) .

The beta--gamma identity,

Γ (1/2) = π

, and the Gamma reflection formula give

\frac{1}{4} B (\frac{1}{4}, \frac{1}{2}) = \frac{Γ ( 1/4 ) π}{4Γ ( 3/4 )}, Γ (1/4) Γ (3/4) = π 2 .

Therefore

\int_{0}^{1} \frac{d x}{1 - x ^{4}} = \frac{Γ ( 1/4 ) ^{2}}{32 π} .

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