Consider further the integral ∫ z [ƒ′(z)/ƒ(z)] dz, where ƒ′(z) = dƒ(z)/dz taken round the perimeter of the primary parallelogram; the contribution to this arising from two opposite perimeter points such as z and z + ω is of the form −ω ∫ z [ƒ′(z)/ƒ(z)] dz, which, as z increases from z0 to z0 + ω′, gives, if λ denote the generalized logarithm, − ω {λ [ƒ(z0 + ω′)] − λ[ƒ(z0)]}, that is, since ƒ(z0 + ω′) = ƒ(z0), gives 2πiNω, where N is an integer; similarly the result of the integration along the other two opposite sides is of the form 2πiN′ω′, where N′ is an integer. The integral, however, is equal to 2πi times the sum of the residues of zƒ′(z) / ƒ(z) at the poles interior to the parallelogram. For a zero, of order m, of ƒ(z) at z = a, the contribution to this sum is 2πima, for a pole of order n at z = b the contribution is −2πinb; we thus infer that Σma − Σnb = Nω + N′ω′; this we express in words by saying that the sum of the values of z where ƒ(z) = 0 within any parallelogram is equal to the sum of the values of z where ƒ(z) = ∞ save for integral multiples of the periods. By considering similarly the function ƒ(z) − A where A is an arbitrary constant, we prove that each of these sums is equal to the sum of the values of z where the function takes the value A in the parallelogram.
We pass now to the construction of a function having two arbitrary periods ω, ω′ of unreal ratio, which has a single pole of the second order in any one of its parallelograms.
For this consider first the network of parallelograms whose corners are the points Ω = mω + m′ω′, where m, m′ take all positive and negative integer values; putting a small circle about each corner of this network, let P be a point outside all these circles; this will be interior to a parallelogram whose corners in order may be denoted by z0, z0 + ω, z0 + ω + ω′, z0 + ω′; we shall denote z0, z0 + ω by A0, B0; this parallelogram Π0 is surrounded by eight other parallelograms, forming with Π0 a larger parallelogram Π1, of which one side, for instance, contains the points z0 − ω − ω′, z0 − ω′, z0 − ω′ + ω, z0 − ω′ + 2ω, which we shall denote by A1, B1, C1, D1. This parallelogram Π1 is surrounded by sixteen of the original parallelograms, forming with Π1 a still larger parallelogram Π2 of which one side, for instance, contains the points z0 − 2ω − 2ω′, z0 − ω − 2ω′, z0 − 2ω′, z0 + ω − 2ω′, z0 + 2ω − 2ω′, z0 + 3ω − 2ω′, which we shall denote by A2, B2, C2, D2, E2, F2. And so on. Now consider the sum of the inverse cubes of the distances of the point P from the corners of all the original parallelograms. The sum will contain the terms
| S0 = | 1 | + ( | 1 | + | 1 | + | 1 | ) + ( | 1 | + | 1 | + ... + | 1 | ) + ... |
| PA03 | PA13 | PB13 | PC13 | PA23 | PB23 | PE23 |
and three other sets of terms, each infinite in number, formed in a similar way. If the perpendiculars from P to the sides A0B0, A1B1C1, A2B2C2D2E2, and so on, be p, p + q, p + 2q and so on, the sum S0 is at most equal to
| 1 | + | 3 | + | 5 | + ... + | 2n + 1 | + ... |
| p3 | (p + q)3 | (p + 2q)3 | (p + nq)3 |
of which the general term is ultimately, when n is large, in a ratio of equality with 2q−3 n−2, so that the series S0 is convergent, as we know the sum Σn−2 to be; this assumes that p ≠ 0; if P be on A0B0 the proof for the convergence of S0 − 1/PA03, is the same. Taking the three other sums analogous to S0 we thus reach the result that the series
φ(z) = −2Σ (z − Ω)−3,
where Ω is mω + m′ω′, and m, m′ are to take all positive and negative integer values, and z is any point outside small circles described with the points Ω as centres, is absolutely convergent. Its sum is therefore independent of the order of its terms. By the nature of the proof, which holds for all positions of z outside the small circles spoken of, the series is also clearly uniformly convergent outside these circles. Each term of the series being a monogenic function of z, the series may therefore be differentiated and integrated outside these circles, and represents a monogenic function. It is clearly periodic with the periods ω, ω′; for φ(z + ω) is the same sum as φ(z) with the terms in a slightly different order. Thus φ(z + ω) = φ(z) and φ(z + ω′) = φ(z).
Consider now the function