 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/7e/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

For

Re z > 0

I (z + 1) = \int_{0}^{\infty} t^{z} e^{- t} d t = [- t^{z} e^{- t}]_{0}^{\infty} + z \int_{0}^{\infty} t^{z - 1} e^{- t} d t .

I (z + 1) = z I (z) .

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