Page:EB1911 - Volume 11.djvu/337

Rh equation ƒ(s, z) = 0. Such an integral ƒK(z, s)dz is called an Abelian Integral.

§ 19. Reversion of an Algebraic Integral.—In a limited number of cases the equation u = ∫ [z0 to z] H(z)dz, in which H(z) is an algebraic function of z, defines z as a single valued function of u. Several cases of this have been mentioned in the previous section; from what was previously proved under § 14, Doubly Periodic Functions, it appears that it is necessary for this that the integral should have at most two linearly independent additive constants of indeterminateness; for instance, for an integral

u = ∫ z z0 [(z − a) (z − b) (z − c) (z − d) (z − e) (z − f) ]−1/2dz,

there are three such constants, of the form A − B, A − C, A − D, which are not connected by any linear equation with integral coefficients, and z is not a single valued function of u.

§ 20. Elliptic Integrals.—An integral of the form ∫ R(z, s)dz, where s denotes the square root of a quartic polynomial in z, which may reduce to a cubic polynomial, and R denotes a rational function of z and s, is called an elliptic integral.

To each value of z belong two values of s, of opposite sign; starting, for some particular value of z, with a definite one of these two values, the sign to be attached to s for any other value of z will be determined by the path of integration for z. When z is in the neighbourhood of any finite value z0 for which the radical s is not zero, if we put z − z0 = t, we can find s − s0 = a power series in t, say s=s0 + Q(t); when z is in the neighbourhood of a value, a, for which s vanishes, if we put z = a + t2, we shall obtain s = tQ(t), where Q(t) is a power series in t; when z is very large and s2 is a quartic polynomial in z, if we put z−1 = t, we shall find s−1 = t2Q(t); when z is very large and s2 is a cubic polynomial in z, if we put z−1 = t2, we shall find s −l = t3Q(t). By means of substitutions of these forms the character of the integral ∫ R(z, s)dz may be investigated for any position of z; in any case it takes a form ∫ [Ht −m + Kt −m+1 + ... + Pt−1 + R + St + ... ]dt involving only a finite number of negative powers of t in the subject of integration. Consider first the particular case ∫ s−1dz; it is easily seen that neither for any finite nor for infinite values of z can negative powers of t enter; the integral is everywhere finite, and is said to be of the first kind; it can, moreover, be shown without difficulty that no integral ∫ R(z, s)dz, save a constant multiple of ∫ s−1dz, has this property. Consider next, s2 being of the form a0z4 + 4a1z3 + ..., wherein a0 may be zero, the integral ∫ (a0z2 + 2a1z) s−1dz; for any finite value of z this integral is easily proved to be everywhere finite; but for infinite values of z its value is of the form At−1 + Q(t), where Q(t) is a power series; denoting by √a0 a particular square root of a0 when a0 is not zero, the integral becomes infinite for z = ∞ for both signs of s, the value of A being + √a0 or − √a0 according as s is √a0·z2 (1 + [2a1/a0] z−1 + ... ) or is the negative of this; hence the integral J1 = ∫ ( [a0z2 + 2a1z]/s + √a0) dz becomes infinite when z is infinite, for the former sign of s, its infinite term being 2√a0·t−1 or 2a0·z, but does not become infinite for z infinite for the other sign of s. When a0 = 0 the signs of s for z = ∞ are not separated, being obtained one from the other by a circuit of z about an infinitely large circle, and the form obtained represents an integral becoming infinite as before for z = ∞, its infinite part being 2√a1·t−1 or 2√a1·√z. Similarly if z0 be any finite value of z which is not a root of the polynomial ƒ(z) to which s2 is equal, and s0 denotes a particular one of the determinations of s for z=z0, the integral

wherein ƒ′(z) = dƒ(z)/dz, becomes infinite for z = z0, s = s0, but not for z = z0, s = −s0. its infinite term in the former case being the negative of 2s0(z − z0). For no other finite or infinite value of z is the integral infinite. If z = be a root of ƒ(z), in which case the corresponding value of s is zero, the integral

becomes infinite for z=0, its infinite part being, if z − = t2, equal to −[ƒ′] t−1: and this integral is not elsewhere infinite. In each of these cases, of the integrals J1, J2, J3, the subject of integration has been chosen so that when the integral is written near its point of infinity in the form ∫[At−2 + Bt−1 + Q(t)] dt, the coefficient B is zero, so that the infinity is of algebraic kind, and so that, when there are two signs distinguishable for the critical value of z, the integral becomes infinite for only one of these. An integral having only algebraic infinities, for finite or infinite values of z, is called an integral of the second kind, and it appears that such an integral can be formed with only one such infinity, that is, for an infinity arising only for one particular, and arbitrary, pair of values (s, z) satisfying the equation s2 = ƒ(z), this infinity being of the first order. A function having an algebraic infinity of the mth order (m > 1), only for one sign of s when these signs are separable, at (1) z = ∞, (2) z = z0, (3) z = a, is given respectively by (s d/dz)m−1 J1, (s d/dz)m−1 J2, (s d/dz)m−1 J3, as we easily see. If then we have any elliptic integral having algebraic infinities we can, by subtraction from it of an appropriate sum of constant multiples of J1, J2, J3 and their differential coefficients just written down, obtain, as the result, an integral without algebraic infinities. But, in fact, if J, J1 denote any two of the three integrals J1, J2, J3, there exists an equation AJ + BJ′ + Cƒs−1dz = rational function of s, z, where A, B, C are properly chosen constants. For the rational function

is at once found to become infinite for (z0, s0), not for (z0, −s0), its infinite part for the first point being 2s/(z − z0), and to become infinite for z infinitely large, and one sign of s only when these are separable, its infinite part there being 2z √a0 or 2 √a1 √z when a0 = 0. It does not become infinite for any other pair (z, s) satisfying the relation s2 = ƒ(z); this is in accordance with the easily verified equation

and there exists the analogous equation

Consider now the integral

this is at once found to be infinite, for finite values of z, only for (z0, s0), its infinite part being log (z − z0), and for z = ∞, for one sign of s only when these are separable, its infinite part being −log t, that is −log z when a0 &ne; 0, and −log (zundefined) when a0 = 0. And, if ƒ = 0, the integral

is infinite at z =, s = 0 with an infinite part log t, that is log (z − )undefined, is not infinite for any other finite value of z, and is infinite like P for z = ∞. An integral possessing such logarithmic infinities is said to be of the third kind.

Hence it appears that any elliptic integral, by subtraction from it of an appropriate sum formed with constant multiples of the integral J3 and the rational functions of the form (s d/dz)m−1 J1 with constant multiples of integrals such as P or P1, with constant multiples of the integral u = ∫s−1dz, and with rational functions, can be reduced to an integral H becoming infinite only for z = ∞, for one sign of s only when these are separable, its infinite part being of the form A log t, that is, A log z or A log (zundefined). Such an integral H = ∫R(z, s)dz does not exist, however, as we at once find by writing R(z, s) = P(z) + sQ(z), where P(z), Q(z) are rational functions of z, and examining the forms possible for these in order that the integral may have only the specified infinity. An analogous theorem holds for rational functions of z and s; there exists no rational function which is finite for finite values of z and is infinite only for z = ∞ for one sign of s and to the first order only; but there exists a rational function infinite in all to the first order for each of two or more pairs (z, s), however they may be situated, or infinite to the second order for an arbitrary pair (z, s); and any rational function may be formed by a sum of constant multiples of functions such as

and their differential coefficients.

The consideration of elliptic integrals is therefore reducible to that of the three

respectively of the first, second and third kind. Now the equation s2 = a0z4 + ... = a0 (z − ) (z − ) (z − ) (z − ), by putting

y = 2s (z − )−2 [a0 ( − ) ( − ) ( − ) ]−1/2

is at once reduced to the form y2 = 4x3 − g2x − g3 = 4(x − e1) (x − e2) (x − e3), say; and these equations enable us to express s and z rationally in terms of x and y. It is therefore sufficient to consider three elliptic integrals

Of these consider the first, putting

where the limits involve not only a value for x, but a definite sign for the radical y. When x is very large, if we put x−1 = t2, y−1 = 2t3 (1 − g2t4 −  g3t6)$−1⁄2$, we have

u = ∫ t 0 (1 + g2t4 + ... ) dt = t +  g2t5 + ...,