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The fractional-part continuation gives $ζ (0) = - 1 / 2$

Example

$ζ (0) = - \frac{1}{2} .$

Facts & Assumptions

Given: The fractional-part continuation of the zeta function.

[L1]

For $Re s > 0$ and $s \neq 1$ , $ζ (s) = \frac{s}{s - 1} - s \int_{1}^{\infty} {x} x^{- s - 1} d x$ (For $Re s > 0$ , zeta admits the fractional-part integral formula with a simple residue-one pole at $1$ ).

Here ${x} = x - ⌊ x ⌋$ .

Verification

technique · direct

1.1givenalgebra

Put $q (x) = {x} - 1 / 2$ and $Q (x) = \int_{1}^{x} q (t) d t$ . The function $q$ has integral zero over each interval $[n, n + 1]$ , so $Q$ is bounded. Integration by parts shows that, for $Re s > - 1$ , $\int_{1}^{\infty} q (x) x^{- s - 1} d x = (s + 1) \int_{1}^{\infty} Q (x) x^{- s - 2} d x .$ The integral on the right converges locally uniformly in this half-plane and defines a holomorphic function there.

2.1L1step 1.1algebra∎

For $Re s > 0$ , $\int_{1}^{\infty} {x} x^{- s - 1} d x = \frac{1}{2 s} + \int_{1}^{\infty} q (x) x^{- s - 1} d x .$ Substituting this into [L1] gives $ζ (s) = \frac{s}{s - 1} - \frac{1}{2} - s \int_{1}^{\infty} q (x) x^{- s - 1} d x .$ By step 1.1, the right-hand side extends holomorphically across $s = 0$ and has value $- 1 / 2$ there. Uniqueness of analytic continuation identifies this value with $ζ (0)$ .

Depends on

For $\operatorname{Re}s>0$, zeta admits the fractional-part integral formula with a simple residue-one pole at $1$

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 11 §3 (standard reference, not scraped)

The fractional-part continuation gives ζ(0)=−1/2ζ(0)=−1/2