Let this expression be valid for 0 < |u| < R, and the function defined thereby, which has a pole of the second order for u=0, be denoted by φ(u). In the range in question it is single valued and satisfies the differential equation
[φ′(u)]2 = 4[φ(u)]3 − g2φ(u) − g3;
in terms of it we can write x = φ(u), y = − φ′(u), and, φ′(u) being an odd function, the sign attached to y in the original integral for x = ∞ is immaterial. Now for any two values u, v in the range in question consider the function
| F(u, v) = ¼ [ | φ′(u) − φ′(v) | ]2 − φ(u) − φ(v); |
| φ(u) − φ(v) |
it is at once seen, from the differential equation, to be such that ∂F/∂u = ∂F/∂v; it is therefore a function of u + v; supposing |u + v| < R we infer therefore, by putting v = 0, that
| φ(u + v) = ¼ [ | φ′(u) − φ′(v) | ]2 − φ(u) − φ(v). |
| φ(u) − φ(v) |
By repetition of this equation we infer that if u1, ... un be any arguments each of which is in absolute value less than R, whose sum is also in absolute value less than R, then φ(u1 + ... + un) is a rational function of the 2n functions φ(us), φ′(us); and hence, if |u| < R, that
| φ(u) = H [ φ ( | u | ), φ′ ( | u | ) ], |
| n | n |
where H is some rational function of the arguments φ(u/n), φ′(u/n). In fact, however, so long as |u/n| < R, each of the functions φ(u/n), φ′(u/n) is single valued and without singularity save for the pole at u=0; and a rational function of single valued functions, each of which has no singularities other than poles in a certain region, is also a single valued function without singularities other than poles in this region. We infer, therefore, that the function of u expressed by H [φ(u/n), φ′(u/n)] is single valued and without singularities other than poles so long as |u| < nR; it agrees with φ(u) when |u| < R, and hence furnishes a continuation of this function over the extended range |u| < nR. Moreover, from the method of its derivation, it satisfies the differential equation [φ′(u)]2 = 4[φ(u)]3 − g2φ(u) − g3. This equation has therefore one solution which is a single valued monogenic function with no singularities other than poles for any finite part of the plane, having in particular for u = 0, a pole of the second order; and the method adopted for obtaining this near u=0 shows that the differential equation has no other such solution. This, however, is not the only solution which is a single valued meromorphic function, a the functions φ(u + α), wherein α is arbitrary, being such. Taking now any range of values of u, from u = 0, and putting for any value of u, x = φ(u), y = −φ′(u), so that y2=4x3-g2x-g3, we clearly have
| u = ∫ (∞)(x, y) | dx | ; |
| y |