The inverse of an incomplete elliptic integral of the first kind in Legendre normal form defines the Jacobi elliptic sine. The corresponding inverse of an incomplete elliptic integral of the second kind does not define a known special function: even Wolfram is looking for such an inverse. This presentation will develop expansions for both kinds of incomplete integrals in closed form using partial Bell polynomials, providing rather compact forms for both inverse integrals.

The incomplete integrals of the first and second kind in Legendre normal form will be denoted

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

$E\left(x\right|m)=\int dx\sqrt{\frac{1-m{x}^2}{1-x^2}}$

Both integrals can be explicitly evaluated in terms of Appell functions, which are hypergeometric functions of two arguments. The evaluation requires a generalization of the binomial formula to arbitrary exponents,

$\begin{array}{l}(1-x{)}^a=1-ax+\frac{a(a-1)}{2!}{x}^2-\frac{a(a-1)(a-2)}{3!}{x}^3+\cdots \\ \phantom{(1-x{)}^a}=\sum _{k=0}^{\infty }\frac{\Gamma (-a+k)}{\Gamma (-a)}\frac{x^k}{k!}\end{array}$

and proceeds as follows:

$\begin{array}{l}\int dx(1-x^2{)}^{-1/2}(1-m{x}^2{)}^{\mp 1/2}\\ =\int dx\sum _{p,r=0}^{\infty }\frac{\Gamma (\frac{1}{2}+p)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\pm \frac{1}{2}+r)}{\Gamma (\pm \frac{1}{2})}\frac{m^r}{p!r!}{x}^{2p+2r}\\ =\sum _{p,r=0}^{\infty }\frac{\Gamma (\frac{1}{2}+p)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\pm \frac{1}{2}+r)}{\Gamma (\pm \frac{1}{2})}\frac{\frac{1}{2}}{p+r+\frac{1}{2}}\frac{m^r}{p!r!}{x}^{2p+2r+1}\\ =x\sum _{p,r=0}^{\infty }\frac{\Gamma (\frac{1}{2}+p)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\pm \frac{1}{2}+r)}{\Gamma (\pm \frac{1}{2})}\frac{\Gamma (\frac{1}{2}+p+r)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\frac{1}{2}+1)}{\Gamma (\frac{1}{2}+1+p+r)}\frac{(x^2{)}^p}{p!}\frac{(m{x}^2{)}^r}{r!}\\ =x{F}_1(\frac{1}{2};\frac{1}{2},\pm \frac{1}{2};\frac{3}{2};x^2,m{x}^2)=u\end{array}$

An independent variable for the inversion has been included in the final equality. The only difference between the two cases is a negative sign on the third argument.

These integrals can be inverted with Lagrange inversion: for a function $u=f\left(x\right)$ , the inverse function expanded about a center c is

$x=f^{-1}\left(u\right)=c+\sum _{k=1}^{\infty }\frac{[u-f(c){]}^k}{k!}\underset{x\to c}{lim}\frac{d^{k-1}}{d{x}^{k-1}}[\frac{x-c}{f\left(x\right)-f\left(c\right)}{]}^k$

Since both integrals are zero at the origin, the expansion about that center simplifies to

$x=\sum _{k=1}^{\infty }\frac{u^k}{k!}\underset{x\to 0}{lim}\frac{d^{k-1}}{d{x}^{k-1}}[\frac{1}{F_1(\frac{1}{2};\frac{1}{2},\pm \frac{1}{2};\frac{3}{2};x^2,m{x}^2)}{]}^k$

The inverse power of the Appell functions can be expressed using partial Bell polynomials. They are defined by the generating function

$\frac{1}{k!}(\sum _{p=1}^{\infty }{x}_{}\frac{t^p}{p!}{)}^k=\sum _{n=k}^{\infty }{B}_{n,k}(x_1,x_2,\dots ,x_{n-k+1})\frac{t^n}{n!}$

The coefficients appearing in the partial polynomials can be evaluated by expansion of the generating function and collection of terms. They can be more efficiently generated recursively with the relation

$B_{n,k}=\sum _{i=1}^{n-k+1}\left(\genfrac{}{}{0ex}{}{n-1}{i-1}\right)x_i{B}_{n-i,k-1}$

The first few nonzero explicit forms are

$\begin{array}{l}{B}_{0,0}=1\\ B_{1,1}=x_1\\ B_{2,1}=x_2\\ B_{2,2}=x_1^2\end{array}\begin{array}{l}{B}_{3,1}=x_3\\ B_{3,2}=3x_1{x}_2\\ B_{3,3}=x_1^3\end{array}\begin{array}{l}{B}_{4,1}=x_4\\ B_{4,2}=3x_2^2+4x_1{x}_3\\ B_{4,3}=6x_1^2{x}_2\\ B_{4,4}=x_1^4\end{array}$

While the partial polynomials are not trivial to generate from scratch, they are known mathematical entities that can be used to express the expansions desired in compact forms.

Partial Bell polynomials are useful in this context because they describe function composition: given two functions expressed as power series,

$f\left(x\right)=\sum _{n=1}^{\infty }{a}_n\frac{x^n}{n!}g\left(x\right)=\sum _{n=1}^{\infty }{b}_n\frac{x^n}{n!}$

the composite function $g\left[f\right(x\left)\right]$ can be expressed as

$g\left[f\right(x\left)\right]=\sum _{n=1}^{\infty }\frac{x^n}{n!}\sum _{k=1}^n{b}_k{B}_{n,k}(a_1,a_2,\dots ,a_{n-k+1})$

Partial Bell polynomials are essentially a bookkeeping method for tracking the combinations appearing in an arbitrary power of an infinite series. The exact same coefficients appear when taking multiple derivatives of a composite function, so that an arbitrary derivative of the composite function $g\left[f\right(x\left)\right]$ can be expressed compactly with partial Bell polynomials in derivatives of $f\left(x\right)$ .

Setting $n=p+r$ and $i=\frac{p-r}{2}$ , the summation for the Appell functions can be written

$F_1=1+\sum _{n=1}^{\infty }\frac{x^{2n}}{2n+1}\frac{1}{n!}\sum _{i=-n/2}^{n/2}\left(\genfrac{}{}{0ex}{}{n}{\frac{n}{2}+i}\right)\frac{\Gamma (\frac{n}{2}+i+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\frac{n}{2}-i\pm \frac{1}{2})}{\Gamma (\pm \frac{1}{2})}{m}^{n/2-i}$

and one can then take

$f\left(x\right)=\sum _{n=1}^{\infty }{a}_n\frac{(x^2{)}^n}{n!}{a}_n=\frac{1}{2n+1}\sum _{i=-n/2}^{n/2}\left(\genfrac{}{}{0ex}{}{n}{\frac{n}{2}+i}\right)\frac{\Gamma (\frac{n}{2}+i+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\frac{n}{2}-i\pm \frac{1}{2})}{\Gamma (\pm \frac{1}{2})}{m}^{n/2-i}$

The expansion of the inverse power is

$(\frac{1}{1+x}{)}^k=\sum _{n=0}^{\infty }(-1{)}^n\frac{\Gamma (k+n)}{\Gamma \left(k\right)}{x}^n=1+g\left(x\right)$

so that for the second part of the composition one can take

$g\left(x\right)=\sum _{n=1}^{\infty }{b}_n\frac{x^n}{n!}{b}_n=(-1{)}^n\frac{\Gamma (k+n)}{\Gamma \left(k\right)}$

The composite functions can now be expressed compactly with partial Bell polynomials:

$[\frac{1}{F_1(\frac{1}{2};\frac{1}{2},\pm \frac{1}{2};\frac{3}{2};x^2,m{x}^2)}{]}^k=1+\sum _{n=1}^{\infty }\frac{(x^2{)}^n}{n!}\sum _{l=1}^n{b}_l{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1})$

The multiple derivative is easily evaluated:

$\begin{array}{l}\frac{d^{k-1}}{d{x}^{k-1}}[\frac{1}{F_1(\frac{1}{2};\frac{1}{2},\pm \frac{1}{2};\frac{3}{2};x^2,m{x}^2)}{]}^k\\ =\sum _{2n\ge k-1}\frac{\left(2n\right)!}{(2n-k+1)!}\frac{x^{2n-k+1}}{n!}\sum _{l=1}^n{b}_l{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1}),k>1\end{array}$

The limit $x\to 0$ is now trivial: merely picking out terms for which

$2n-k+1=0\to k=2n+1$

Application of this restriction to the coefficients $b_l$ can be captured by adding an additional index to these coefficients.

A complete expansion about the origin for both cases can thus be expressed as

$\begin{array}{l}x=u+\sum _{n=1}^{\infty }\frac{u^{2n+1}}{(2n+1)n!}\sum _{l=1}^n{b}_{n,l}{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1})\\ \\ a_n=\frac{1}{2n+1}\sum _{i=-n/2}^{n/2}\left(\genfrac{}{}{0ex}{}{n}{\frac{n}{2}+i}\right)\frac{\Gamma (\frac{n}{2}+i+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\frac{n}{2}-i\pm \frac{1}{2})}{\Gamma (\pm \frac{1}{2})}{m}^{n/2-i}\\ \\ b_{n,l}=(-1{)}^l\frac{\Gamma (2n+l+1)}{\Gamma (2n+1)}\end{array}$

It is now straightforward to evaluate the first few terms of this series in Mathematica or the like. Taking the upper signs this gives

$\begin{array}{l}u-\frac{1+m}{6}{u}^3+\frac{1+14m+m^2}{120}{u}^5-\frac{1+135m+135m^2+m^3}{5040}{u}^7\\ +\frac{1+1228m+5478m^2+1228m^3+m^4}{362880}{u}^9-\dots \end{array}$

which precisely matches the expansion of the Jacobi elliptic sine about the origin. Since as stated at the outset this function inverts the elliptic integral of the first kind, the match confirms the accuracy of the method. That means that taking lower signs will give the first few terms of the inverse of the elliptic integral of the second kind:

$\begin{array}{l}u-\frac{1-m}{6}{u}^3+\frac{1-14m+13m^2}{120}{u}^5-\frac{1-135m+627m^2-493m^3}{5040}{u}^7\\ +\frac{1-1228m+19662m^2-55804m^3+37369m^4}{362880}{u}^9-\dots \end{array}$

The numerical coefficients of each polynomial in m are symmetric for the inverse elliptic integral of the first kind, but not for that of the second kind.

In summary, the inverse of an incomplete elliptic integral of the first kind can be expressed compactly as

$\begin{array}{l}{F}^{-1}\left(u\right|m)=u+\sum _{n=1}^{\infty }\frac{u^{2n+1}}{(2n+1)n!}\sum _{l=1}^n{b}_{n,l}{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1})\\ \\ a_n=\frac{1}{2n+1}\sum _{i=-n/2}^{n/2}\left(\genfrac{}{}{0ex}{}{n}{\frac{n}{2}+i}\right)\frac{\Gamma (\frac{n}{2}+i+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\frac{n}{2}-i+\frac{1}{2})}{\Gamma (+\frac{1}{2})}{m}^{n/2-i}\\ \\ b_{n,l}=(-1{)}^l\frac{\Gamma (2n+l+1)}{\Gamma (2n+1)}\end{array}$

while the inverse of the incomplete integral of the second kind can be expressed compactly as

$\begin{array}{l}{E}^{-1}\left(u\right|m)=u+\sum _{n=1}^{\infty }\frac{u^{2n+1}}{(2n+1)n!}\sum _{l=1}^n{b}_{n,l}{B}_{n,l}(a_1,a_2,\dots ,a_{n-l+1})\\ \\ a_n=\frac{1}{2n+1}\sum _{i=-n/2}^{n/2}\left(\genfrac{}{}{0ex}{}{n}{\frac{n}{2}+i}\right)\frac{\Gamma (\frac{n}{2}+i+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma (\frac{n}{2}-i-\frac{1}{2})}{\Gamma (-\frac{1}{2})}{m}^{n/2-i}\\ \\ b_{n,l}=(-1{)}^l\frac{\Gamma (2n+l+1)}{\Gamma (2n+1)}\end{array}$

The radii of convergence for these series will be controlled by the singularities of the Appell functions. Since these functions converge as long as both arguments of the independent variable are less than unity, a rough estimate for the radii of convergence can be taken to be $\frac{1}{\sqrt{m}}$ .

These expressions appear to be more compact and concise than available existing expansions. They can be evaluated numerically by any library that handles partial Bell polynomials, such as Math.

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Uploaded 2019.05.27 — Updated 2019.05.29 analyticphysics.com