The Jacobi zeta function can be defined in terms of incomplete and complete elliptic integrals as

$Z\left(\phi \right|m)=E\left(\phi \right|m)-\frac{E\left(m\right)}{K\left(m\right)}F\left(\phi \right|m)$

where Jacobi amplitude appears explicitly, and the integrals themselves are

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

The elliptic parameter m is used for simplicity rather than the elliptic modulus k. The two are related by $m=k^2$ .

To derive Fourier series for the constituent integrals of the zeta function, begin with a generalization of the binomial formula to arbitrary exponents:

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

The integrands of the incomplete integrals expanded with this generalization are

$\begin{array}{l}\frac{1}{\sqrt{1-m{sin}^2\phi }}=\sum _{n=0}^{\infty }\frac{\Gamma (\frac{1}{2}+n)}{\Gamma \left(\frac{1}{2}\right)}\frac{m^n}{n!}{sin}^{2n}\phi \\ \sqrt{1-m{sin}^2\phi }=\sum _{n=0}^{\infty }\frac{\Gamma (-\frac{1}{2}+n)}{\Gamma (-\frac{1}{2})}\frac{m^n}{n!}{sin}^{2n}\phi \end{array}$

and differ only in the signs on the fractions. Using the standard binomial theorem, an arbitrary even power of the sine function can be rewritten as

$\begin{array}{l}{\sin }^{2n}x=\frac{1}{(2i{)}^{2n}}(e^{ix}-e^{-ix}{)}^{2n}\\ \phantom{{\sin }^{2n}x}=\frac{(-1{)}^n}{2^{2n}}\sum _{p=0}^{2n}(-1{)}^p\left(\genfrac{}{}{0ex}{}{2n}{p}\right)e^{i(2n-p)x}{e}^{-ipx}\\ {\sin }^{2n}x=\frac{1}{2^{2n}}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)+\frac{1}{2^{2n-1}}\sum _{p=0}^{n-1}(-1{)}^{n+p}\left(\genfrac{}{}{0ex}{}{2n}{p}\right)\cos 2(n-p)x\end{array}$

For later convenience let the summation in the second term run over the combination $n-p$ :

${\sin }^{2n}x=\frac{1}{2^{2n}}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)+\frac{1}{2^{2n-1}}\sum _{k=1}^n(-1{)}^k\left(\genfrac{}{}{0ex}{}{2n}{n-k}\right)\cos 2kx$

When this expression is inserted into the integrands and a single integration performed, the incomplete integrals will be expressed as Fourier sine series.

Since integration of a cosine function with even argument between zero and $\frac{\pi }{2}$ is identically zero, the constant terms before integration are proportional to the complete elliptic integrals. First simplify the binomial coefficient using the Legendre duplication formula,

$\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=\frac{\Gamma (2n+1)}{\Gamma (n+1{)}^2}=2^{2n}\frac{\Gamma (n+\frac{1}{2})}{\Gamma \left(\frac{1}{2}\right)\Gamma (n+1)}$

so that the constant term for the incomplete integral of the first kind is

$\sum _{n=0}^{\infty }\frac{\Gamma (\frac{1}{2}+n{)}^2}{\Gamma (\frac{1}{2}{)}^2}\frac{\Gamma \left(1\right)}{\Gamma (1+n)}\frac{m^n}{n!}={}_{2}F_1(\frac{1}{2},\frac{1}{2};1;m)\equiv \frac{2}{\pi }K\left(m\right)$

and that for the incomplete integral of the second kind is

$\sum _{n=0}^{\infty }\frac{\Gamma (-\frac{1}{2}+n)}{\Gamma (-\frac{1}{2})}\frac{\Gamma (\frac{1}{2}+n)}{\Gamma \left(\frac{1}{2}\right)}\frac{\Gamma \left(1\right)}{\Gamma (1+n)}\frac{m^n}{n!}={}_{2}F_1(-\frac{1}{2},\frac{1}{2};1;m)\equiv \frac{2}{\pi }E\left(m\right)$

The binomial coefficients for the remaining terms can be expressed using the Legendre duplication formula as

$\left(\genfrac{}{}{0ex}{}{2n}{n-k}\right)=\frac{\Gamma (2n+1)}{\Gamma (n-k+1)\Gamma (n+k+1)}=2^{2n}\frac{\Gamma (n+\frac{1}{2})\Gamma (n+1)}{\Gamma \left(\frac{1}{2}\right)\Gamma (n-k+1)\Gamma (n+k+1)}$

The summation over the remaining terms for both incomplete integrals can now be rearranged by interchanging the order of summation. Since the sums for $n=0$ are empty, the remaining terms are

$\begin{array}{l}\sum _{n=1}^{\infty }\frac{\Gamma (\pm \frac{1}{2}+n)}{\Gamma (\pm \frac{1}{2})}{m}^n\sum _{k=1}^{n}2(-1{)}^k\frac{\Gamma (\frac{1}{2}+n)}{\Gamma \left(\frac{1}{2}\right)\Gamma (n-k+1)\Gamma (n+k+1)}cos2k\phi \\ =\sum _{k=1}^{\infty }2(-1{)}^{k}cos2k\phi \sum _{n=k}^{\infty }\frac{\Gamma (\pm \frac{1}{2}+n)}{\Gamma (\pm \frac{1}{2})}\frac{\Gamma (\frac{1}{2}+n)}{\Gamma \left(\frac{1}{2}\right)\Gamma (n-k+1)\Gamma (n+k+1)}{m}^n\\ =\sum _{k=1}^{\infty }2(-1{)}^{k}cos2k\phi \sum _{p=0}^{\infty }\frac{\Gamma (\pm \frac{1}{2}+k+p)}{\Gamma (\pm \frac{1}{2})}\frac{\Gamma (\frac{1}{2}+k+p)}{\Gamma \left(\frac{1}{2}\right)\Gamma (2k+1+p)}\frac{m^{k+p}}{p!}\\ =\sum _{k=1}^{\infty }\frac{(-1{)}^k}{2^{2k-1}}\frac{\Gamma (k\pm \frac{1}{2})}{\Gamma (\pm \frac{1}{2})}\frac{m^k}{k!}{}_{2}F_1(k\pm \frac{1}{2},k+\frac{1}{2};2k+1;m)cos2k\phi \\ =\sum _{k=1}^{\infty }\frac{(-m{)}^k}{2^{2k-1}}\left(\genfrac{}{}{0ex}{}{k-1\pm \frac{1}{2}}{k}\right){}_{2}F_1(k\pm \frac{1}{2},k+\frac{1}{2};2k+1;m)cos2k\phi \end{array}$

where the upper signs are for the integral of the first kind and the lower signs for the second. The hypergeometric function here can be rewritten as a linear combination of complete elliptic integrals with rational coefficients, but since there does not appear to be a closed form for the coefficients it is more efficient to leave it as is.

After a single integration with respect to the angular variable, the Fourier series for the incomplete integral of the first kind is thus

$\begin{array}{l}F\left(\phi \right|m)={}_{2}F_1(\frac{1}{2},\frac{1}{2};1;m)\phi \\ +\sum _{k=1}^{\infty }\frac{(-m{)}^k}{2^{2k}k}\left(\genfrac{}{}{0ex}{}{k-\frac{1}{2}}{k}\right){}_{2}F_1(k+\frac{1}{2},k+\frac{1}{2};2k+1;m)sin2k\phi \end{array}$

while that of the second kind is

$\begin{array}{l}E\left(\phi \right|m)={}_{2}F_1(-\frac{1}{2},\frac{1}{2};1;m)\phi \\ +\sum _{k=1}^{\infty }\frac{(-m{)}^k}{2^{2k}k}\left(\genfrac{}{}{0ex}{}{k-\frac{3}{2}}{k}\right){}_{2}F_1(k-\frac{1}{2},k+\frac{1}{2};2k+1;m)sin2k\phi \end{array}$

The linear terms have be written as hypergeometric functions to display their relations to the remaining terms more clearly. These linear terms disappear when the zeta function is formed, with the result

$\begin{array}{l}Z\left(\phi \right|m)=\sum _{k=1}^{\infty }\frac{(-m{)}^k}{2^{2k}k}C(k,m)sin2k\phi \\ \\ C(k,m)=\left(\genfrac{}{}{0ex}{}{k-\frac{3}{2}}{k}\right){}_{2}F_1(k-\frac{1}{2},k+\frac{1}{2};2k+1;m)\\ -\frac{E\left(m\right)}{K\left(m\right)}\left(\genfrac{}{}{0ex}{}{k-\frac{1}{2}}{k}\right){}_{2}F_1(k+\frac{1}{2},k+\frac{1}{2};2k+1;m)\end{array}$

The Fourier series is easily visualized in comparison with the full function using a JavaScript library that supports the Jacobi zeta function, such as Math. Here are the first thirty terms of the series in red overlaid on the full function:

And here is the difference between the full function and the first thirty terms of the series:

The truncated series is quite accurate as long as m < 0.9 or thereabouts.

The motivation behind the development of a Fourier series for the Jacobi zeta function comes from the explicit temporal evolution of the Kepler problem. The Kapteyn series for the y-coordinate in that presentation is

$\frac{y}{a}=\frac{2\sqrt{1-e^2}}{e}\sum _{k=1}^{\infty }\frac{J_k\left(ke\right)}{k}sink\omega t$

When this series is plotted without overall factors, it looks rather suspiciously like a Jacobi zeta function

except that it has half the frequency and leans in the opposite direction for large values of the eccentricity. This can be handled by considering the function $Z\left(\frac{\pi -\phi }{2}\right|m)$ , which when numerically normalized and plotted in green with the Kapteyn series looks like this:

Particularly for small eccentricity, it appears that the zeta function is a good fit to the Kapteyn series for an appropriate choice of elliptic parameter. For large eccentricity the elliptic modulus must approach closer and closer to unity for the chance of a fit, which is not easily presented and so not supported here or in the next graphic.

The Fourier series for the modified Jacobi zeta function is

$Z\left(\frac{\pi -\phi }{2}\right|m)=\sum _{k=1}^{\infty }\frac{(-m{)}^k}{2^{2k}k}C(k,m)sin(k\pi -k\phi )=-\sum _{k=1}^{\infty }\frac{m^k}{2^{2k}k}C(k,m)sink\phi$

so if it is possible for the two series to match for some relation between eccentricity and elliptic parameter one must compare the two quantities

$J_k\left(ke\right)\mathrm{and}-\frac{m^k}{2^{2k}}C(k,m)$

Unfortunately, even after normalizing the right-hand quantity to match the Bessel function for $k=1$ there does not appear to be a choice for the elliptic parameter that produces an exact match between the two quantities:

This simple visual inspection implies that the Jacobi zeta function is a decent approximation to the y-coordinate in the Kepler problem, but cannot be an exact solution. Schade!

Uploaded 2018.09.08 — Updated 2018.09.23 analyticphysics.com