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22 Jacobian Elliptic FunctionsProperties

§22.14 Integrals

Contents
  1. §22.14(i) Indefinite Integrals of Jacobian Elliptic Functions
  2. §22.14(ii) Indefinite Integrals of Powers of Jacobian Elliptic Functions
  3. §22.14(iii) Other Indefinite Integrals
  4. §22.14(iv) Definite Integrals

§22.14(i) Indefinite Integrals of Jacobian Elliptic Functions

With x,

22.14.1 sn(x,k)dx =k1ln(dn(x,k)kcn(x,k)),
22.14.2 cn(x,k)dx =k1Arccos(dn(x,k)),
22.14.3 dn(x,k)dx =Arcsin(sn(x,k))=am(x,k).

The branches of the inverse trigonometric functions are chosen so that they are continuous. See §22.16(i) for am(z,k).

Secondly,

22.14.4 cd(x,k)dx=k1ln(nd(x,k)+ksd(x,k)),
22.14.5 sd(x,k)dx=(kk)1Arcsin(kcd(x,k)),
22.14.6 nd(x,k)dx=k1Arccos(cd(x,k)).

Again, the branches of the inverse trigonometric functions must be continuous.

Lastly, with 0<x<2K,

For alternative, and symmetric, formulations of the results in this subsection see Carlson (2006a).

§22.14(ii) Indefinite Integrals of Powers of Jacobian Elliptic Functions

See §22.16(ii). The indefinite integral of the 3rd power of a Jacobian function can be expressed as an elementary function of Jacobian functions and a product of Jacobian functions. The indefinite integral of a 4th power can be expressed as a complete elliptic integral, a polynomial in Jacobian functions, and the integration variable. See Lawden (1989, pp. 87–88). See also Gradshteyn and Ryzhik (2015, §§5.131–5.134) and Carlson (2006a).

For indefinite integrals of squares and products of even powers of Jacobian functions in terms of symmetric elliptic integrals, see Carlson (2006b).

§22.14(iii) Other Indefinite Integrals

In (22.14.13)–(22.14.15), 0<x<2K.

22.14.13 dxsn(x,k)=ln(sn(x,k)cn(x,k)+dn(x,k)),
22.14.14 cn(x,k)dxsn(x,k)=12ln(1dn(x,k)1+dn(x,k)),
22.14.15 cn(x,k)dxsn2(x,k)=dn(x,k)sn(x,k).

For additional results see Gradshteyn and Ryzhik (2015, §5.13) and Lawden (1989, Chapter 3).

§22.14(iv) Definite Integrals

22.14.16 0K(k)ln(sn(t,k))dt=π4K(k)12K(k)lnk,
22.14.17 0K(k)ln(cn(t,k))dt=π4K(k)+12K(k)ln(k/k),
22.14.18 0K(k)ln(dn(t,k))dt=12K(k)lnk.

Corresponding results for the subsidiary functions follow by subtraction; compare (22.2.10).