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 Incoming links: Sine

Since the odd sine part of the complex exponential integrates to zero,

\frac{1}{2 π} \int_{- n}^{n} e^{ik x} d k = \frac{1}{π} \int_{0}^{n} cos (k x) d k .

Part (e)(ii) therefore yields

n \to \infty lim \frac{1}{2 π} \int_{- n}^{n} e^{ik x} d k = δ (x)

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} .

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