 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-3/8e/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

For

x > 0

, differentiate the absolutely convergent geometric series to obtain

(1 - e^{- x})^{- 2} = \sum_{j = 0}^{\infty} (j + 1) e^{- j x}, (e^{x} - 1)^{- 2} = \sum_{m = 2}^{\infty} (m - 1) e^{- m x} .

σ = ℜ z > 1

\sum_{m \geq 2} (m - 1) \int_{0}^{\infty} x^{σ} e^{- m x} d x = Γ (σ + 1) \sum_{m \geq 2} \frac{m - 1}{m ^{σ + 1}} < \infty.

z

. The substitution

u = m x

\int_{0}^{\infty} \frac{x ^{z}}{( e ^{x} - 1 ) ^{2}} d x = Γ (z + 1) m \geq 2 \sum (m^{- z} - m^{- z - 1}) = Γ (z + 1) (ζ (z) - ζ (z + 1)) .

The missing

m = 1

terms cancel in the difference of the two Riemann zeta functions.

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