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.

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

Jacobi Elliptic Functions of Fixed Period

Quasi-Period of the Complex Jacobi Amplitude Function

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

$F\left(\phi \right|m)=\underset{0}{\overset{\phi}{\int}}\frac{d\phi}{\sqrt{1-m{sin}^{2}\phi}}$

$E\left(\phi \right|m)=\underset{0}{\overset{\phi}{\int}}d\phi \phantom{\rule{.3em}{0ex}}\sqrt{1-m{sin}^{2}\phi}$

$\mathrm{\Pi}(n;\phi |m)=\underset{0}{\overset{\phi}{\int}}\frac{d\phi}{(1-n{sin}^{2}\phi )\sqrt{1-m{sin}^{2}\phi}}$

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 $m={k}^{2}$. There is also traditionally a complementary elliptic modulus defined by ${{k}^{\prime}}^{2}=1-{k}^{2}$, 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=sin\phi $ gives the incomplete elliptic integrals in Legendre normal form:

$F\left(\phi \right|m)=\underset{0}{\overset{sin\phi}{\int}}\frac{dx}{\sqrt{(1-{x}^{2})(1-m{x}^{2})}}$

$E\left(\phi \right|m)=\underset{0}{\overset{sin\phi}{\int}}dx\phantom{\rule{.3em}{0ex}}\sqrt{\frac{1-m{x}^{2}}{1-{x}^{2}}}$

$\mathrm{\Pi}(n;\phi |m)=\underset{0}{\overset{sin\phi}{\int}}\frac{dx}{(1-n{x}^{2})\sqrt{(1-{x}^{2})(1-m{x}^{2})}}$

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 $\frac{\pi}{2}$ defines the complete elliptic integrals:

$K\left(m\right)=\underset{0}{\overset{\frac{\pi}{2}}{\int}}\frac{d\phi}{\sqrt{1-m{sin}^{2}\phi}}$

$E\left(m\right)=\underset{0}{\overset{\frac{\pi}{2}}{\int}}d\phi \phantom{\rule{.3em}{0ex}}\sqrt{1-m{sin}^{2}\phi}$

$\mathrm{\Pi}\left(n\right|m)=\underset{0}{\overset{\frac{\pi}{2}}{\int}}\frac{d\phi}{(1-n{sin}^{2}\phi )\sqrt{1-m{sin}^{2}\phi}}$

A simple angular substitution of the form

$\begin{array}{llll}\phi =\frac{\pi}{2}-\theta & & & sin\phi =cos\theta \\ d\phi =-d\theta & & & cos\phi =sin\theta \end{array}$

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:

$\begin{array}{l}K(-m)=\underset{0}{\overset{\frac{\pi}{2}}{\int}}\frac{d\phi}{\sqrt{1+m{sin}^{2}\phi}}=\underset{0}{\overset{\frac{\pi}{2}}{\int}}\frac{d\phi}{\sqrt{m+1-m{cos}^{2}\phi}}\\ K(-m)=\frac{1}{\sqrt{m+1}}K\left(\frac{m}{m+1}\right)\end{array}$

$\begin{array}{l}E(-m)=\underset{0}{\overset{\frac{\pi}{2}}{\int}}d\phi \phantom{\rule{.3em}{0ex}}\sqrt{1+m{sin}^{2}\phi}=\underset{0}{\overset{\frac{\pi}{2}}{\int}}d\phi \phantom{\rule{.3em}{0ex}}\sqrt{m+1-m{cos}^{2}\phi}\\ E(-m)=\sqrt{m+1}\phantom{\rule{.3em}{0ex}}E\left(\frac{m}{m+1}\right)\end{array}$

$\begin{array}{l}\mathrm{\Pi}\left(n\right|-m)=\underset{0}{\overset{\frac{\pi}{2}}{\int}}\frac{d\phi}{(1-n{sin}^{2}\phi )\sqrt{1+m{sin}^{2}\phi}}\\ \phantom{\mathrm{\Pi}\left(n\right|-m)}=\underset{0}{\overset{\frac{\pi}{2}}{\int}}\frac{d\phi}{(1-n+n{cos}^{2}\phi )\sqrt{m+1-m{cos}^{2}\phi}}\\ \mathrm{\Pi}\left(n\right|-m)=\frac{1}{(1-n)\sqrt{m+1}}\mathrm{\Pi}\left(\frac{n}{n-1}\right|\frac{m}{m+1})\end{array}$

With an argument of $\frac{1}{m}$ 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 $\frac{\pi}{2}$ , split the integral into two parts:

$K\left(\frac{1}{m}\right)=\underset{0}{\overset{{sin}^{-1}\sqrt{m}}{\int}}\frac{d\phi}{\sqrt{1-\frac{1}{m}{sin}^{2}\phi}}-i\underset{{sin}^{-1}\sqrt{m}}{\overset{\frac{\pi}{2}}{\int}}\frac{d\phi}{\sqrt{\frac{1}{m}{sin}^{2}\phi -1}}$

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

$sin\phi =\sqrt{m}sin\alpha \phantom{\rule{4em}{0ex}}cos\phi \phantom{\rule{.3em}{0ex}}d\phi =\sqrt{m}cos\alpha \phantom{\rule{.3em}{0ex}}d\alpha $

under which the integral becomes

$\underset{0}{\overset{{sin}^{-1}\sqrt{m}}{\int}}\frac{d\phi}{\sqrt{1-\frac{1}{m}{sin}^{2}\phi}}=\underset{0}{\overset{\frac{\pi}{2}}{\int}}\frac{\sqrt{m}cos\alpha \phantom{\rule{.3em}{0ex}}d\alpha}{\sqrt{1-{sin}^{2}\alpha}\sqrt{1-m{sin}^{2}\alpha}}=\sqrt{m}K\left(m\right)$

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

$cos\phi =\sqrt{1-m}cos\alpha \phantom{\rule{4em}{0ex}}sin\phi \phantom{\rule{.3em}{0ex}}d\phi =\sqrt{1-m}sin\alpha \phantom{\rule{.3em}{0ex}}d\alpha $

under which the integral becomes

$\underset{{sin}^{-1}\sqrt{m}}{\overset{\frac{\pi}{2}}{\int}}\frac{d\phi}{\sqrt{\frac{1}{m}{sin}^{2}\phi -1}}=\underset{0}{\overset{\frac{\pi}{2}}{\int}}\frac{\sqrt{1-m}sin\alpha \phantom{\rule{.3em}{0ex}}d\alpha}{\sqrt{\frac{1-(1-m){cos}^{2}\alpha}{m}-1}\sqrt{1-(1-m){cos}^{2}\alpha}}=\sqrt{m}K(1-m)$

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

$K\left(\frac{1}{m}\right)=\sqrt{m}\left[K\right(m)-iK(1-m\left)\right]$

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

$K\left(m\right)=\frac{1}{\sqrt{m}}\left[K\right(\frac{1}{m})-iK(1-\frac{1}{m}\left)\right]$

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

The elliptic nome is defined by

$q\left(m\right)=exp[-\pi \frac{K(1-m)}{K\left(m\right)}]$

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

$\frac{K(1+m)}{K(-m)}=\frac{\sqrt{m+1}}{K\left(\frac{m}{m+1}\right)}\times \frac{1}{\sqrt{m+1}}\left[K\right(\frac{1}{m+1})-iK(1-\frac{1}{m+1}\left)\right]=\frac{K\left(\frac{1}{m+1}\right)}{K\left(\frac{m}{m+1}\right)}-i$

so that the elliptic nome for real negative argument is

$q(-m)=-exp[-\pi \frac{K\left(\frac{1}{m+1}\right)}{K\left(\frac{m}{m+1}\right)}]$

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

$\begin{array}{l}\frac{K(1-m)}{K\left(m\right)}=\frac{K[-(m-1\left)\right]}{K\left(m\right)}\\ \phantom{\frac{K(1-m)}{K\left(m\right)}}=\frac{K\left(\frac{m-1}{m}\right)}{\sqrt{m}}\times \frac{\sqrt{m}}{K\left(\frac{1}{m}\right)-iK(1-\frac{1}{m})}\\ \frac{K(1-m)}{K\left(m\right)}=\frac{K(1-\frac{1}{m})}{K(\frac{1}{m}{)}^{2}+K(1-\frac{1}{m}{)}^{2}}\left[K\right(\frac{1}{m})+iK(1-\frac{1}{m}\left)\right]\end{array}$

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

$q\left(m\right)=exp\left[\pi \frac{K(1-\frac{1}{m})}{K(\frac{1}{m}{)}^{2}+K(1-\frac{1}{m}{)}^{2}}\left[K\right(\frac{1}{m})+iK(1-\frac{1}{m}\left)\right]\right]$

is no longer entirely real in this domain.

Consider cases for which the differential equation

$\frac{dx}{\sqrt{f\left(x\right)}}=\frac{dy}{\sqrt{f\left(y\right)}}$

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

$g(x,y)[x\sqrt{f\left(y\right)}-y\sqrt{f\left(x\right)}]$

and form its differential:

$\begin{array}{l}g[\sqrt{f\left(y\right)}dx-\sqrt{f\left(x\right)}dy+\frac{x{f}^{\prime}\left(y\right)}{2\sqrt{f\left(y\right)}}dy-\frac{y{f}^{\prime}\left(x\right)}{2\sqrt{f\left(x\right)}}dx]\\ \phantom{\rule{4em}{0ex}}+[x\sqrt{f\left(y\right)}-y\sqrt{f\left(x\right)}][{g}_{x}dx+{g}_{y}dy]\\ \phantom{\rule{2em}{0ex}}=g\sqrt{f\left(x\right)f\left(y\right)}[\frac{dx}{\sqrt{f\left(x\right)}}-\frac{dy}{\sqrt{f\left(y\right)}}]\\ \phantom{\rule{4em}{0ex}}+\frac{dx}{\sqrt{f\left(x\right)}}[{g}_{x}x\sqrt{f\left(x\right)f\left(y\right)}-{g}_{x}yf\left(x\right)-\frac{y}{2}g{f}^{\prime}\left(x\right)]\\ \phantom{\rule{4em}{0ex}}-\frac{dy}{\sqrt{f\left(y\right)}}[{g}_{y}y\sqrt{f\left(x\right)f\left(y\right)}-{g}_{y}xf\left(y\right)-\frac{x}{2}g{f}^{\prime}\left(y\right)]\end{array}$

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

$g(x,y)=g\left(xy\right)$

under which the differential becomes

$\begin{array}{l}[g+{g}^{\prime}xy]\sqrt{f\left(x\right)f\left(y\right)}[\frac{dx}{\sqrt{f\left(x\right)}}-\frac{dy}{\sqrt{f\left(y\right)}}]\\ \phantom{\rule{2em}{0ex}}-\frac{dx}{\sqrt{f\left(x\right)}}[{g}^{\prime}{y}^{2}f\left(x\right)+\frac{y}{2}g{f}^{\prime}\left(x\right)]\\ \phantom{\rule{2em}{0ex}}+\frac{dy}{\sqrt{f\left(y\right)}}[{g}^{\prime}{x}^{2}f\left(y\right)+\frac{x}{2}g{f}^{\prime}\left(y\right)]\end{array}$

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

${g}^{\prime}{y}^{2}f\left(x\right)+\frac{y}{2}g{f}^{\prime}\left(x\right)=h(x,y)={g}^{\prime}{x}^{2}f\left(y\right)+\frac{x}{2}g{f}^{\prime}\left(y\right)$

then the differential becomes

$[g+{g}^{\prime}xy-h]\sqrt{f\left(x\right)f\left(y\right)}[\frac{dx}{\sqrt{f\left(x\right)}}-\frac{dy}{\sqrt{f\left(y\right)}}]$

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

$\frac{{g}^{\prime}}{g}=\frac{1}{2}\frac{y{f}^{\prime}\left(x\right)-x{f}^{\prime}\left(y\right)}{{x}^{2}f\left(y\right)-{y}^{2}f\left(x\right)}$

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 $f\left(x\right)=1-{x}^{2}$ , for which

$\frac{{g}^{\prime}}{g}=0\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}g\left(xy\right)=1$

The solution in this case to the differential equation is

$x\sqrt{1-{y}^{2}}-y\sqrt{1-{x}^{2}}=c$

Direct integration of the differential equation gives

$\begin{array}{c}\underset{0}{\overset{x}{\int}}\frac{dt}{\sqrt{1-{t}^{2}}}-\underset{0}{\overset{y}{\int}}\frac{dt}{\sqrt{1-{t}^{2}}}=\underset{0}{\overset{c}{\int}}\frac{dt}{\sqrt{1-{t}^{2}}}\\ \\ {sin}^{-1}x-{sin}^{-1}y={sin}^{-1}c\end{array}$

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

$x=sinu\phantom{\rule{5em}{0ex}}y=sinv$

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 $f\left(x\right)=1-{x}^{4}$ that arose in the context of arc length on a lemniscate. The condition for an integrable solution is

$\frac{{g}^{\prime}}{g}=-\frac{2xy}{1+{x}^{2}{y}^{2}}\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}g\left(xy\right)=\frac{1}{1+{x}^{2}{y}^{2}}$

so that the solution to the differential equation is

$\frac{x\sqrt{1-{y}^{4}}-y\sqrt{1-{x}^{4}}}{1+{x}^{2}{y}^{2}}=c$

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 $m=-1$ ,

$x=sn\left(u\right|-1)\phantom{\rule{5em}{0ex}}y=sn\left(v\right|-1)$

this solution represents the addition formula for those functions:

$\frac{snucnvdnv-snvcnudnu}{1+{sn}^{2}u{sn}^{2}v}=sn(u-v)$

The final easily integrable case is $f\left(x\right)=(1-{x}^{2})(1-m{x}^{2})$ for elliptic integrals of the first kind. The condition for integrability is

$\frac{{g}^{\prime}}{g}=\frac{2mxy}{1-m{x}^{2}{y}^{2}}\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}g\left(xy\right)=\frac{1}{1-m{x}^{2}{y}^{2}}$

so that the solution to the differential equation is

$\frac{x\sqrt{(1-{y}^{2})(1-m{y}^{2})}-y\sqrt{(1-{x}^{2})(1-m{x}^{2})}}{1-m{x}^{2}{y}^{2}}=c$

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

$x=sn\left(u\right|m)\phantom{\rule{5em}{0ex}}y=sn\left(v\right|m)$

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

$\frac{snucnvdnv-snvcnudnu}{1-m{sn}^{2}u{sn}^{2}v}=sn(u-v)$

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.

Basic derivatives of Jacobi elliptic functions and relations among them:

$\begin{array}{l}\frac{d}{du}snu=cnudnu\\ \frac{d}{du}cnu=-snudnu\\ \frac{d}{du}dnu=-msnucnu\end{array}\phantom{\rule{3em}{0ex}}\begin{array}{r}{sn}^{2}u+{cn}^{2}u=1\\ {dn}^{2}u+m{sn}^{2}u=1\end{array}$

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:

$\begin{array}{c}\frac{d}{du}snu=\sqrt{1-{sn}^{2}u}\sqrt{1-m{sn}^{2}u}\\ x=snu\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}u=\int \frac{dx}{\sqrt{(1-{x}^{2})(1-m{x}^{2})}}\end{array}$

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:

${sn}^{-1}x=\underset{0}{\overset{x}{\int}}\frac{dx}{\sqrt{(1-{x}^{2})(1-m{x}^{2})}}$

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.

${cn}^{-1}x=\underset{x}{\overset{1}{\int}}\frac{dx}{\sqrt{(1-{x}^{2})(1-m+m{x}^{2})}}$

${dn}^{-1}x=\underset{x}{\overset{1}{\int}}\frac{dx}{\sqrt{(1-{x}^{2})(m-1+{x}^{2})}}$

${ns}^{-1}x=\underset{x}{\overset{\infty}{\int}}\frac{dx}{\sqrt{({x}^{2}-1)({x}^{2}-m)}}$

${nc}^{-1}x=\underset{1}{\overset{x}{\int}}\frac{dx}{\sqrt{({x}^{2}-1)[m+(1-m\left){x}^{2}\right]}}$

${nd}^{-1}x=\underset{1}{\overset{x}{\int}}\frac{dx}{\sqrt{({x}^{2}-1)[1+(m-1\left){x}^{2}\right]}}$

${sc}^{-1}x=\underset{0}{\overset{x}{\int}}\frac{dx}{\sqrt{(1+{x}^{2})[1+(1-m\left){x}^{2}\right]}}$

${sd}^{-1}x=\underset{0}{\overset{x}{\int}}\frac{dx}{\sqrt{(1+m{x}^{2})[1+(m-1\left){x}^{2}\right]}}$

${cs}^{-1}x=\underset{x}{\overset{\infty}{\int}}\frac{dx}{\sqrt{(1+{x}^{2})(1-m+{x}^{2})}}$

${cd}^{-1}x=\underset{x}{\overset{1}{\int}}\frac{dx}{\sqrt{(1-{x}^{2})(1-m{x}^{2})}}$

${ds}^{-1}x=\underset{x}{\overset{\infty}{\int}}\frac{dx}{\sqrt{(m+{x}^{2})(m-1+{x}^{2})}}$

${dc}^{-1}x=\underset{1}{\overset{x}{\int}}\frac{dx}{\sqrt{({x}^{2}-1)({x}^{2}-m)}}$

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.

The Jacobi elliptic functions of a real variable have periods proportional to complete elliptic integrals of the first kind,

$\begin{array}{l}sn[t+4K(m),m]=sn(t,m)\\ cn[t+4K(m),m]=cn(t,m)\\ dn[t+2K(m),m]=dn(t,m)\end{array}$

where the elliptic parameter has been written explicitly for the elliptic functions. If one is trying to match an elliptic function to an unknown function with a given period different from these periods, then a frequency multiplying the first argument is generally necessary. This alters the resulting function in a nonlinear fashion, unlike the corresponding change to a circular function, even though the function retains its period in terms of elliptic integrals.

The alteration can be understood by comparing two elliptic functions for different frequencies and parameters:

$\begin{array}{l}sn[{\omega}_{1}t+4K({m}_{1}),{m}_{1}]=sn\left[{\omega}_{1}\right(t+\frac{4K\left({m}_{1}\right)}{{\omega}_{1}}),{m}_{1}]\\ sn[{\omega}_{2}t+4K({m}_{2}),{m}_{2}]=sn\left[{\omega}_{2}\right(t+\frac{4K\left({m}_{2}\right)}{{\omega}_{2}}),{m}_{2}]\end{array}$

In order for these two functions to have the same period, the quantities added to the independent variable must be equal:

$\frac{K\left({m}_{1}\right)}{{\omega}_{1}}=\frac{K\left({m}_{2}\right)}{{\omega}_{2}}$

Starting from one frequency and parameter and specifying a new frequency, the parameter that maintains the original fixed period is given by

$K\left({m}_{2}\right)=\frac{{\omega}_{2}}{{\omega}_{1}}K\left({m}_{1}\right)\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}{m}_{2}={K}^{-1}\left[\frac{{\omega}_{2}}{{\omega}_{1}}K\left({m}_{1}\right)\right]$

While the Jacobi elliptic sine formally inverts an elliptic integral of the first kind, in practice the inversion must be done with numerical root finding for the simple reason that the elliptic function itself depends on the parameter to be determined.

The technique can be graphically illustrated with a Jacobi elliptic sine constrained to a period of 2π:

Alternately, starting from an elliptic parameter of zero with the frequency corresponding to the known fixed period, changes to the elliptic parameter can be accommodated by adjusting the frequency to

${\omega}_{2}=\frac{K\left({m}_{2}\right)}{K\left({m}_{1}\right)}{\omega}_{1}$

which can again be graphically illustrated with a Jacobi elliptic sine constrained to a period of 2π:

Plotting the function $sn[\frac{K\left(m\right)}{K\left(0\right)}t,m]$ along with a sine function and allowing variation of the elliptic parameter, both positive and negative, nicely illustrates the analytic relationship between the two functions:

The behavior of this last graphic for negative values of the parameter indicates that there is an interesting way to generalize the circular function behavior

$sin(t+\frac{\pi}{2})=cost$

to Jacobi elliptic functions with fixed period. Consider the relationship

$sn[\frac{K\left(n\right)}{K\left(0\right)}(t+\frac{\pi}{2}),n]\stackrel{?}{=}cn[\frac{K\left(m\right)}{K\left(0\right)}t,m]$

Numerical rooting finding for some nontrivial argument, say $t=\frac{\pi}{4}$ , indicates that one can indeed locate values for the elliptic parameters that make the two functions equivalent. This is demonstrated by the following interactive graphic:

The default red line is a Jacobi sine function with
$n=0$ ,
*i.e.* a regular sine function. Checking the box locates the value of *n*, shown rounded to eight decimal places, that makes the two functions equivalent. Note that this value is negative relative to the other elliptic parameter.

Evaluating both sides of the relationship under consideration with a CAS, for example Mathematica, and equating their leading terms gives

$K\left(m\right)=\sqrt{1-n}\phantom{\rule{.3em}{0ex}}K\left(n\right)$

as the analytic relationship connecting the two Jacobi elliptic functions with fixed period. The right-hand side can be rewritten using an identity above for negative parameter,

$\sqrt{1-n}\phantom{\rule{.3em}{0ex}}K\left(n\right)=K\left(\frac{n}{n-1}\right)$

which explains why so many simple rational values turn up in the numerical evaluations of the interactive graphic.

To understand this quantity, begin with those of the elliptic integrals. For all of these integrals, the sine function only appears squared. As a result adding a real value of $\pi $ to the argument of the integral of the first kind leads to an additional multiple of the corresponding complete integral,

$\begin{array}{l}F(\phi +\pi |m)=\underset{0}{\overset{\phi +\pi}{\int}}\frac{d\phi}{\sqrt{1-m{sin}^{2}\phi}}\\ \phantom{F(\phi +\pi |m)}=K\left(m\right)+\underset{\frac{\pi}{2}}{\overset{\pi}{\int}}\frac{d\phi}{\sqrt{1-m{sin}^{2}\phi}}+\underset{\pi}{\overset{\phi +\pi}{\int}}\frac{d\phi}{\sqrt{1-m{sin}^{2}\phi}}\\ F(\phi +\pi |m)=2K\left(m\right)+F\left(\phi \right|m)\end{array}$

where the variable of integration is replaced with $\pi -\phi $ in the second intermediate term and by $\phi -\pi $ in the third. Repeated application gives the general rule

$F(\phi +\pi p|m)=F\left(\phi \right|m)+2pK\left(m\right)$

for integral *p*. The rules for integrals of the second and third kind are mutatis mutandis identical:

$E(\phi +\pi p|m)=E\left(\phi \right|m)+2pE\left(m\right)$

$\mathrm{\Pi}(\phi +\pi p|m)=\mathrm{\Pi}(n,\phi |m)+2p\mathrm{\Pi}\left(n\right|m)$

These rules can be interpreted to hold over the entire complex plane for both the argument and the parameter.

The Jacobi amplitude function is defined as the inverse of the elliptic integral of the first kind:

$u=\underset{0}{\overset{\phi}{\int}}\frac{d\phi}{\sqrt{1-m{sin}^{2}\phi}}\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}\phi =am(u,m)$

Adding twice the complete integral to each side of this equation, reversing the calculation above, gives

$u+2K\left(m\right)=\underset{0}{\overset{\phi +\pi}{\int}}\frac{d\phi}{\sqrt{1-m{sin}^{2}\phi}}$

so that for values of the elliptic parameter that keep the complete elliptic integral real, once can apply this repeatedly for the rule

$am[u+2pK(m\left)\right|m]=am\left(u\right|m)+\pi p$

While this looks as simple to apply as the rules for elliptic integrals, in practice there are some catches. The Jacobi elliptic functions have an second period in $iK(1-m)$ . When evaluating the amplitude function as the inverse sine of a Jacobi elliptic sine, any real part of this second period will change the overall sign of the inverse sine, but does not alter the result otherwise. Additionally, any imaginary part of the elliptic parameter negates the entire rule, so that only the inverse sine is then necessary.

These last comments remain to be fully understodd but are stated here to document how the function is evaluated in Math. This is at least how the function appears to be evaluted in Mathematica.

*Uploaded 2017.11.20 — Updated 2018.11.13*
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