 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-3/7e/c/solution

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

For any

z \in / {0, - 1, - 2, \dots}

, choose

N

with

Re (z + N) > 0

B (p, z) = \frac{( p + z ) ( p + z + 1 ) \dots ( p + z + N - 1 )}{z ( z + 1 ) \dots ( z + N - 1 )} B (p, z + N) .

The recurrence makes this independent of

N

and agrees with the original integral where

Re z > 0

. It is the desired analytic continuation away from the listed points.

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