Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 7B (Further Complex Methods)

Table of contents
- a 7B
  - Solution a
- b 7B
  - Solution b

a

Solution

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.

b

Solution

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 .

Use the sum-and-ratio substitution for gamma integrals

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

whose Jacobian is

r

. Absolute convergence for

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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