58.1. PERIODIC FUNCTIONS 1827

Now from the claim,

∑w̸=0

1

|w|3

= ∑(m,n)̸=(0,0)

1

|mw1 +nw2|3≤ ∑

(m,n)̸=(0,0)

1

k3 (|m|+ |n|)3

=1k3

∞

∑j=1

∑|m|+|n|= j

1

(|m|+ |n|)3 =1k3

∞

∑j=1

4 jj3 < ∞.

Now consider the series in 58.1.6. Letting z ∈ B(0,R) ,

℘(z) ≡ 1z2 + ∑

w̸=0,|w|≤R

(1

(z−w)2 −1

w2

)

+ ∑w̸=0,|w|>R

(1

(z−w)2 −1

w2

)and the last series converges uniformly on B(0,R) to an analytic function. Thus ℘ is ameromorphic function and also the argument given above involving differentiation of theseries termwise is valid. Thus ℘ is an elliptic function as claimed. This is called theWeierstrass ℘ function. This has proved the following theorem.

Theorem 58.1.13 The function ℘ defined above is an example of an elliptic function. Onany compact set, ℘ equals a rational function added to a series which is uniformly andabsolutely convergent on the compact set.

58.1.3 The Differential Equation Satisfied By ℘

For z not a pole,

℘′ (z) =

−2z3 − ∑

w̸=0

2

(z−w)3

Also since there are no poles of order 1 you can obtain a primitive for ℘, −ζ .2 To doso, recall

℘(z)≡ 1z2 + ∑

w ̸=0

(1

(z−w)2 −1

w2

)where for |z|< R this is the sum of a rational function with a uniformly convergent series.Therefore, you can take the integral along any path, γ (0,z) from 0 to z which misses thepoles of ℘. By the uniform convergence of the above integral, you can interchange the sumwith the integral and obtain

ζ (z) =1z+ ∑

w̸=0

1z−w

+z

w2 +1w

(58.1.7)

2I don’t know why it is traditional to refer to this antiderivative as −ζ rather than ζ but I am following theconvention. I think it is to minimize the number of minus signs in the next expression.