This page will document useful transformations of elliptic integrals and relations among them, along with their derivations. These can be applied to simplifications of analytic formulae or numerical evaluations.

Contents

Definitions of Integrals
Complete Integrals of Negative Parameter
Reciprocal Modulus Transformations
Elliptic Nome on the Real Axis
Integrable Cases and Addition Formulae
Inverse Jacobi Elliptic Functions


Definitions of Integrals

The incomplete elliptic integrals of the first, second and third kinds in trigonometric form are

F(φ|m) =0φ dφ 1-m sin2φ

E(φ|m) =0φ dφ 1-m sin2φ

Π(n; φ|m) =0φ dφ (1-n sin2φ) 1-m sin2φ

with 0<m<1 in these defining integrals. The elliptic parameter m will be used on this page rather than the traditional elliptic modulus k for two reasons:

1) The resulting formulae are generally cleaner and simpler.

2) Computer algebra systems generally use the parameter rather than the modulus. Having formulae in the same form reduces unintended errors in application.

The elliptic parameter and modulus are related by m=k2. There is also traditionally a complementary elliptic modulus defined by k2 =1-k2, and functions of this complementary modulus are often denoted with a prime.

While there is a variety of mathematical notation for the elliptic integrals, it appears to be common to distinguish those using the parameter with a vertical bar before that argument, while those using the modulus are denoted with a comma before that argument.

The transformation x=sinφ gives the incomplete elliptic integrals in Legendre normal form:

F(φ|m) =0 sinφ dx (1-x2) (1-mx2 )

E(φ|m) =0 sinφ dx 1-mx2 1-x2

Π(n; φ|m) =0 sinφ dx (1-n x2) (1-x2) (1-mx2 )

The angular variable is left as an argument on the left-hand sides to aid in avoiding errors in application.

Setting the angular argument equal to π2 defines the complete elliptic integrals:

K(m) =0 π2 dφ 1-m sin2φ

E(m) =0 π2 dφ 1-m sin2φ

Π(n|m) =0 π2 dφ (1-n sin2φ) 1-m sin2φ


Complete Integrals of Negative Parameter

A simple angular substitution of the form

φ=π2-θ sinφ=cosθ dφ=dθ cosφ=sinθ

interchanges sines and cosines while reversing the order of integration. This implies that the complete integrals can be equally defined using sines or cosines, since the integrals are otherwise identical after this substitution.

That fact can be used to evaluate the complete integrals for negative parameter quite easily:

K(m) =0 π2 dφ 1+m sin2φ =0 π2 dφ m+1-m cos2φ K(m) =1m+1 K(m m+1)


E(m) =0 π2 dφ 1+m sin2φ =0 π2 dφ m+1-m cos2φ E(m) =m+1 E(m m+1)


Π(n| m) =0 π2 dφ (1-n sin2φ) 1+m sin2φ Π(n| m) =0 π2 dφ (1-n+n cos2φ) m+1-m cos2φ Π(n| m) =1 (1-n) m+1 Π(n n-1| mm+1)


Reciprocal Modulus Transformations

With an argument of 1m the denominator of the complete integral of the first kind acquires a zero. Since the real part of the inverse sine function is always less than π2 , split the integral into two parts:

K(1m) =0 sin1 m dφ1 -1m sin2φ -i sin1 m π2 dφ1m sin2φ-1

These two integrals can be converted to complete elliptic integrals with trigonometric substitutions that preserve the valid endpoint while scaling the other to a value appropriate for a complete integral. For the integral in the real part, the substitution that preserves the lower endpoint is

sinφ=msinα cosφdφ =mcosα dα

under which the integral becomes

0 sin1 m dφ1 -1m sin2φ =0 π2 mcosα dα 1-sin2α 1-msin2 α =mK(m)

For the integral in the imaginary part, the substitution that preserves the upper endpoint is

cosφ=1-m cosα sinφdφ =1-msinα dα

under which the integral becomes

sin1 m π2 dφ1m sin2φ-1 =0 π2 1-msinα dα 1-(1-m) cos2αm -1 1-(1-m) cos2α =mK (1-m)

since the complete integrals can be defined with either sines or cosines as noted above. The reciprocal modulus transformation for the complete elliptic integral of the first kind is thus

K(1m) =m[ K(m) -iK( 1-m)]

This result technically only holds for Imm>0 : for Imm<0 the imaginary part of the result requires a change in sign.

For m>1 this relation can be used to separate the real and imaginary parts of the complete integral when written in the form

K(m) =1m[ K(1m) -iK( 1-1m)]

since then the arguments on the right-hand side are both within the defining range of convergence.


Elliptic Nome on the Real Axis

The elliptic nome is defined by

q(m) =exp[π K(1-m) K(m)]

For 0<m<1 the arguments of the elliptic integrals are within the defining range, so that both integrals as well as the nome are real.

For m<0 the arguments of both integrals are outside the defining range. The complete integral with negative parameter has been given above, and the integral with argument greater than one can be rewritten using the reciprocal modulus transformation. The ratio of the two integrals is

K(1+m) K(m) =m+1 K(m m+1) ×1m+1 [K(1 m+1) -iK(1-1 m+1)] =K(1 m+1) K(m m+1) -i

so that the elliptic nome for real negative argument is

q(m) =exp[π K(1 m+1) K(m m+1) ]

and since both arguments are now within the defining range, the nome is purely real.

For m>1 the arguments of both integrals are again outside the defining range but with their respective roles reversed. The ratio of the two integrals is now

K(1-m) K(m) =K[ (m-1)] K(m) K(1-m) K(m) =K(m-1 m)m ×m K(1m) -iK(1 -1m) K(1-m) K(m) =K(1 -1m) K(1m )2 +K(1 -1m )2 [K(1m) +iK(1 -1m)]

The arguments on the right-hand side are now within the defining range, so that this ratio is explicitly separated into real and imaginary parts. The nome itself

q(m) =exp[π K(1 -1m) K(1m )2 +K(1 -1m )2 [K(1m) +iK(1 -1m)]]

is no longer entirely real in this domain.


Integrable Cases and Addition Formulae

Consider cases for which the differential equation

dx f(x) =dy f(y)

is readily integrable, leading to an addition formula for functions inverting an integral. Start with a solution of the form

g(x,y) [xf(y) -yf(x)]

and form its differential:

g[f(y) dx -f(x) dy +x f(y) 2f(y) dy -y f(x) 2f(x) dx] +[xf(y) -yf(x)] [gxdx +gydy] =gf(x) f(y) [dx f(x) -dy f(y)] +dx f(x) [gxx f(x) f(y) -gxy f(x) -y2g f (x)] -dy f(y) [gyy f(x) f(y) -gyx f(y) -x2g f (y)]

The first bracketed term is zero by construction. The first term in each of the two other brackets can be made equal by choosing the overall function multiplier to have the dependence

g(x,y) =g(xy)

under which the differential becomes

[g +gxy] f(x) f(y) [dx f(x) -dy f(y)] -dx f(x) [g y2 f(x) +y2g f (x)] +dy f(y) [g x2 f(y) +x2g f (y)]

where the prime on g represents a derivative with respect to its entire argument. If the last two bracketed quantities can be made equal,

g y2 f(x) +y2g f (x) =h(x,y) =g x2 f(y) +x2g f (y)

then the differential becomes

[g +gxy -h] f(x) f(y) [dx f(x) -dy f(y)]

and an integrable solution will have been found that is a constant linking the function under the radical at two different argument values. The condition for equality can be written

gg =12 y f(x) -xf (y) x2 f(y) -y2 f(x)

so that whenever the right-hand combination can be written as a function of the product xy the differential equation is exactly integrable.


For explicit solutions, start with the motivating case f(x) =1-x2  , for which

gg =0 g(xy) =1

The solution in this case to the differential equation is

x1-y2 -y1-x2 =c

Direct integration of the differential equation gives

0x dt 1-t2 -0y dt 1-t2 =0c dt 1-t2   sin1x -sin1y =sin1c

and taking a sine of both sides reproduces the solution. Replacing the variables in the solution with functions that invert the integrals

x=sinu y=sinv

the solution to the differential equation becomes

sinucosv -sinvcosu =c =sin(u-v)

so that the solution to the differential equation represents the addition formula for the sine function, albeit with one negative angle.

Next consider the historically interesting case f(x) =1-x4 that arose in the context of arc length on a lemniscate. The condition for an integrable solution is

gg =2xy 1+x2 y2 g(xy) =11+x2 y2

so that the solution to the differential equation is

x1 -y4 -y1 -x4 1+x2 y2 =c

This is one way to express what is known as Euler’s algebraic addition theorem for elliptic integrals.

Recognizing that the integrals in this case are inverted by Jacobi elliptic functions with m=1 ,

x=sn (u|1) y=sn (v|1)

this solution represents the addition formula for those functions:

snucnvdnv -snvcnudnu 1+sn2u sn2v =sn(u-v)

The final easily integrable case is f(x) =(1-x2) (1-mx2) for elliptic integrals of the first kind. The condition for integrability is

gg =2mxy 1-mx2 y2 g(xy) =11 -mx2 y2

so that the solution to the differential equation is

x (1-y2) (1-my2) -y (1-x2) (1-mx2) 1-mx2 y2 =c

Since the integrals in this case are inverted by Jacobi elliptic functions of arbitrary parameter,

x=sn (u|m) y=sn (v|m)

this solution represents the general addition formula for Jacobi elliptic functions:

snucnvdnv -snvcnudnu 1-msn2u sn2v =sn(u-v)

This is by far a simpler and more direct derivation of the addition formula for Jacobi elliptic functions than the one given in Whittaker & Watson!


Having a general condition for testing the integrability of the differential equation, at least for a solution of the given form, it is straightforward to examine other rational expressions for possible integrability. Unfortunately there do not appear to be any other such simple expressions. In particular the rational expression defining elliptic integrals of the second kind is not integrable in this form.

Here is a nice historical overview by Jose Barrios for some context to this section.


Inverse Jacobi Elliptic Functions

Basic derivatives of Jacobi elliptic functions and relations among them:

ddusnu =cnudnu dducnu =snudnu ddudnu =msnucnu sn2u +cn2u =1 dn2u +msn2u =1

It is well known that the incomplete elliptic integral of the first kind is inverted by the Jacobi sn function. This is demonstrated by rewriting the derivative of that function with the relations:

ddusnu =1-sn2u 1-msn2u x=snu u=dx (1-x2) (1-mx2)

Since the result of the integral is the argument of the elliptic function, it and the elliptic function are inverses of each other. This provides an integral representation of the inverse Jacobi sn function:

sn1x =0x dx (1-x2) (1-mx2)

The same process can be carried out for the remaining eleven Jacobi elliptic functions, giving an integral representation of each. Limits of integration are chosen to keep the functions real for suitable choices of argument.

cn1x =x1 dx (1-x2) (1-m +mx2)

dn1x =x1 dx (1-x2) (m-1 +x2)

ns1x =x dx (x2-1) (x2-m)

nc1x =1x dx (x2-1) [m+(1-m) x2]

nd1x =1x dx (x2-1) [1+(m-1) x2]

sc1x =0x dx (1+x2) [1+(1-m) x2]

sd1x =0x dx (1+mx2) [1+(m-1) x2]

cs1x =x dx (1+x2) (1-m+x2)

cd1x =x1 dx (1-x2) (1-mx2)

ds1x =x dx (m+x2) (m-1 +x2)

dc1x =1x dx (x2-1) (x2-m)

It is rather remarkable that all of these inverse functions have similar integral forms, all of which are incomplete elliptical integrals of the first kind or their complex continuations.


Uploaded 2017.11.20 — Updated 2017.11.30 analyticphysics.com