Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 150 / 2 / c

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- Solution c

Solution

Write the second form of the functional equation as

ζ (s) = χ (s) ζ (1 - s), χ (s) = 2^{s} π^{s - 1} sin (π s /2) Γ (1 - s) .

For

- 2 \leq σ \leq 2

, the elementary exponential formula for the sine gives, uniformly for

t \geq 4

,

∣ sin (π s /2) ∣ ≍ e^{π t /2} .

The stated Stirling formula gives

∣Γ (1 - s) ∣ ≍ t^{1/2 - σ} e^{- π t /2}

uniformly on the same strip. The bounded factors

2^{σ} π^{σ - 1}

and the cancelling exponentials therefore show that

∣ χ (s) ∣ ≍ t^{1/2 - σ} .

Taking absolute values in the functional equation proves the Vertical-strip factor in the Riemann zeta functional equation:

∣ ζ (s) ∣ ≍ t^{1/2 - σ} ∣ ζ (1 - s) ∣ .

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