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The Basel sum is pi squared over six by a residue computation

Statement

$\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6} .$

This computation uses $π \cot (π z) / z^{2}$ directly. It does not follow by substituting $f (z) = 1 / z^{2}$ into the cotangent summation theorem, because that theorem excludes integer poles of $f$ .

Facts & Assumptions

Given: The meromorphic function $F (z) = π \cot (π z) / z^{2}$ .

[L1]

The zeros of $\sin (π z)$ are the integers and are simple (The zeros of complex sine are the integer multiples of pi, and the zeros of complex cosine are the odd half-integer multiples of pi, Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives).

[L2]

The residue theorem applies on expanding rectangles, and the same boundary estimate as in the cotangent summation proof makes the rectangle integral of $F$ tend to $0$ (The residue theorem for a null-homologous cycle).

Proof

technique · direct

1.1

At every nonzero integer $n$ , the function $π \cot (π z)$ has residue $1$ , so $F$ has residue $1 / n^{2}$ there.

L1algebra

1.2

Near $0$ one has $\sin (π z) = π z - \frac{π^{3} z^{3}}{6} + O (z^{5}), \cos (π z) = 1 - \frac{π^{2} z^{2}}{2} + O (z^{4}),$ so $π \cot (π z) = \frac{1}{z} - \frac{π^{2}}{3} z + O (z^{3}) .$ Therefore $F (z) = \frac{1}{z^{3}} - \frac{π^{2}}{3 z} + O (z),$ and the residue of $F$ at $0$ is $- π^{2} / 3$ .

L1algebra

2.1

Integrate $F$ around the rectangles used in the cotangent theorem. By [L2], the boundary integral tends to $0$ , so the sum of the enclosed residues tends to $0$ . Hence $2 \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - \frac{π^{2}}{3} = 0,$ which rearranges to the Basel value.

step 1.1step 1.2L2∎

Depends on

Used by

The residue theorem gives the Basel sum Example

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §5.3 (standard reference, not scraped)