Calculation of the Pm,s Coefficients

We need only \(m=0\) and \(m=\pm2\). Clearly:

\[\begin{split}p_{m,s}&=&\frac{a^3}{2\pi}\int_0^{2\pi/\omega} \frac{e^{-im\Delta \phi(t)}}{r^3(t)}e^{i s \omega t}dt\\ &=& a^3\int_0^{2\pi/\omega} e^{-im\phi_0}\frac{\cos(m\phi(t))-i\sin(m\phi(t))}{r^3(t)} e^{i s \omega t}dt\end{split}\]

For \(m=0\):

\[\begin{split}p_{0,s}&=& \frac{1}{2\pi}\int_{0}^{2\pi} \frac{e^{i s (u-e\sin u)}}{\omega (1-e\cos u)^2} du\\ &=& \frac{1}{2\pi\omega} \int_{0}^{2\pi} \frac{e^{i s (u-e\sin u)}} {(1-e\cos u)^2} du\end{split}\]

From \(1/(1-x)^2=\sum_{k=0}^\infty (k+1)x^k\):

\[p_{0,s}= \sum_{k=0}^\infty \frac{(k+1)e^k}{2\pi\omega} \int_{0}^{2\pi} e^{i s (u-e\sin u)} \cos^k u du\]

Which for \(s=0\), using \(\int_0^{2\pi} \cos^{2k} u du = \frac{2\pi (2k)!}{2^{2k}(k!)^2}\) gives:

\[p_{0,0}=\frac{1}{\omega} \sum_{k=0}^\infty \frac{(2k+1)!}{2^{2k}(k!)^2}e^k\]

And for \(s \neq 0\):

\[\begin{split}p_{0,s} & = & \sum_{k=0}^\infty \frac{(k+1)e^k}{2\pi\omega} \int_{0}^{2\pi} e^{i s (u-e\sin u)} \cos^k u du\\ & = & \sum_{k=0}^\infty \frac{(k+1)e^k}{2^{k+1}\pi\omega} \int_{0}^{2\pi} e^{i s (u-e\sin u)} \left(e^{iu}+e^{-iu}\right)^k du\\ & = & \sum_{k=0}^\infty \frac{(k+1)e^k}{2^{k+1}\pi\omega}\sum_{c=0}^k {k \choose c} \int_{0}^{2\pi} e^{i s (u-e\sin u)} e^{icu}e^{-i(k-c)u} du\\ & = & \sum_{k=0}^\infty \frac{(k+1)e^k}{2^{k+1}\pi\omega}\sum_{c=0}^k {k \choose c} \int_{0}^{2\pi} e^{i (s+2c-k) u} e^{-ies\sin u} du\end{split}\]

If we change variable \(u=u'-\pi/2\Rightarrow \sin u = -\cos u'\):

\[\begin{split}2\pi p_{0,s}&=&\sum_{k=0}^\infty \frac{(k+1)e^k}{2^k\omega} \sum_{c=0}^k {k \choose c} e^{-i (s+2c-k)\pi/2} \int_{\pi/2}^{5\pi/2} e^{i (s+2c-k) u'} e^{ies\cos u'} du'\\ &=&\sum_{k=0}^\infty \frac{(k+1)e^k}{2^k\omega} \sum_{c=0}^k {k \choose c} (-i)^{s+2c-k} \left\{ \int_{\pi/2}^{2\pi} e^{i (s+2c-k) u'} e^{ies\cos u'} du' + \int_{2\pi}^{5\pi/2} e^{i (s+2c-k) u'} e^{ies\cos u'} du' \right\}\\ &=&\sum_{k=0}^\infty \frac{(k+1)e^k}{2^k\omega} \sum_{c=0}^k {k \choose c} (-i)^{s+2c-k} \left\{ \int_{\pi/2}^{2\pi} e^{i (s+2c-k) u'} e^{ies\cos u'} du' + \int_{0}^{\pi/2} e^{i (s+2c-k) u'} e^{ies\cos u'} du' \right\}\\ &=&\sum_{k=0}^\infty \frac{(k+1)e^k}{2^k\omega} \sum_{c=0}^k {k \choose c} (-i)^{s+2c-k} \int_{0}^{2\pi} e^{i (s+2c-k) u'} e^{ies\cos u'} du'\\ &=&\sum_{k=0}^\infty \frac{2\pi(k+1)e^k}{2^k\omega} \sum_{c=0}^k {k \choose c} (-i)^{s+2c-k} i^{s+2c-k} J_{s+2c-k}(es)\\ &=&\sum_{k=0}^\infty \frac{2\pi(k+1)e^k}{2^k\omega} \sum_{c=0}^k {k \choose c} J_{s+2c-k}(es)\\ &=&\sum_{k=0}^\infty \frac{2\pi(k+1)(se)^k}{2^k\omega s^k} \sum_{c=0}^k {k \choose c} \sum_{\lambda=max(0,k-s-2c)}^{\infty} \frac{(-1)^\lambda (se)^{2\lambda+s+2c-k}} {2^{2\lambda+s+2c-k}\lambda!(\lambda+s+2c-k)!}\\ &=&\sum_{k=0}^\infty \frac{2\pi(k+1)(se)^k}{2^k\omega s^k} \sum_{c=0}^k {k \choose c} \sum_{\lambda=max(0,k-s-2c)}^{\infty} \frac{(-1)^\lambda (se)^{2\lambda+s+2c-k}} {2^{2\lambda+s+2c-k}\lambda!(\lambda+s+2c-k)!}\\ &=&\frac{2\pi}{\omega} \left(\frac{es}{2}\right)^s \sum_{k=0}^\infty \frac{(k+1)}{s^k} \sum_{c=0}^k {k \choose c} \sum_{\lambda=max(0,k-s-2c)}^{\infty} \frac{(-1)^\lambda (s^2e^2/4)^{\lambda+c}} {\lambda!(\lambda+s+2c-k)!}\\\end{split}\]

If we now change indices to \(n=\lambda+c\Rightarrow \lambda=n-c,\quad\lambda+s+2c-k=n+s+c-k\). The lower limit on \(\lambda\) gives: \(k-s-2c\leq n-c\Rightarrow c\geq k-n-s\). Finally, in order for the range of \(c\) to not be empty: \(k-n-s\leq n\Rightarrow k\leq 2n+s\). With all these we can write:

\[\begin{split}p_{0,s}=\sum_{n=0}^\infty \alpha_{s,n}\left\{ \begin{array}{l@{,\quad}l} e^{2n} & s=0\\ (se/2)^{s+2n} & s \neq 0 \end{array}\right.\end{split}\]

with

\[\begin{split}\alpha_{s,n}\equiv\frac{1}{\omega}\left\{\begin{array}{l@{,\quad}l} \frac{(2n+1)!}{2^{2n}(n!)^2} & s=0\\ (-1)^n \sum_{k=0}^{2n+s} \frac{k+1}{s^k} \sum_{c=max(0,k-n-s)}^{min(n,k)} {k \choose c} \frac{(-1)^c}{(n-c)!(n+s+c-k)!} & s \neq 0 \end{array} \right.\end{split}\]

Verified using Mathematica.

For \(m=\pm2\) we need:

\[\begin{split}\cos2\phi &=& 1-2\sin^2\phi = 1-2\frac{(1-e^2)\sin^2u}{(1-e\cos u)^2} = 1-\frac{(1-e^2)(1-\cos2u)}{(1-e\cos u)^2}\\ \sin2\phi &=& 2\sin\phi\cos\phi = 2\sqrt{1-e^2} \frac{\sin u(\cos u - e)}{(1-e\cos u)^2} = 2\sqrt{1-e^2}\frac{\sin 2u - e\sin u}{(1-e\cos u)^2}\end{split}\]

Plugging into the expression for \(p_{\pm2,s}\):

\[\begin{split}p_{\pm2,s}&=& \frac{1}{2\pi}\int_0^{2\pi/\omega} a^3\exp(\mp 2i\phi_0)\frac{\cos(2\phi(t))\mp i\sin(2\phi(t))}{r^3(t)} \exp[i s (u-e\sin u)]dt\\ &=& \frac{e^{\mp 2i\phi_0}}{2\pi\omega}\int_0^{2\pi} \left[1-\frac{(1-e^2)(1-\cos2u)}{(1-e\cos u)^2} \mp i\sqrt{1-e^2}\frac{\sin 2u - 2e\sin u}{(1-e\cos u)^2}\right] \frac{\exp[i s (u-e\sin u)]}{(1-e\cos u)^2} du\end{split}\]

Thus we need to evaluate 5 different integrals, the first of which was already done while calculating \(p_{0,s}\):

\[\begin{split}p_{\pm2,s}&=&\exp\left(\mp 2i\phi_0\right) p_{0,s} -\\ &&{}-\frac{\exp(\mp 2i\phi_0)(1-e^2)}{2\pi\omega} \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]} {(1-e\cos u)^4} du+\\ &&{}+\frac{\exp(\mp 2i\phi_0)(1-e^2)}{2\pi\omega} \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]\cos 2u} {(1-e\cos u)^4} du\mp\\ &&{}\mp i\frac{\exp(\mp 2i\phi_0)\sqrt{1-e^2}}{2\pi\omega} \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]\sin 2u} {(1-e\cos u)^4} du\pm\\ &&{}\pm i\frac{2e\exp(\mp 2i\phi_0)\sqrt{1-e^2}}{2\pi\omega} \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]\sin u} {(1-e\cos u)^4} du\\\end{split}\]

To solve them we will use \(1/(1-x)^4=\sum_{k=0}^\infty {{k+3} \choose 3} x^k\) and we will directly calculate the following general integral:

\[\begin{split}2\pi\omega I_{\lambda,s}&\equiv& \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]\exp(i\lambda u)} {(1-e\cos u)^4} du\\ &=&\sum_{k=0}^\infty {{k+3} \choose 3} e^k \int_{0}^{2\pi} \exp[i (s+\lambda) u]\exp[-ise\sin u)]\cos^k u du\\ &=&\sum_{k=0}^\infty {{k+3} \choose 3} \left(\frac{e}{2}\right)^k \sum_{c=0}^k {k \choose c} \int_{0}^{2\pi} \exp[i (s+\lambda+2c-k) u]\exp(-ise\sin u) du\\ &=&\sum_{k=0}^\infty {{k+3} \choose 3} 2\pi \left(\frac{e}{2}\right)^k \sum_{c=0}^k {k \choose c} J_{s+\lambda+2c-k}(es)\\ &=&\sum_{k=0}^\infty {{k+3} \choose 3} 2\pi \left(\frac{e}{2}\right)^k \sum_{c=0}^k {k \choose c} \sum_{\nu=max(0,k-s-\lambda-2c)}^{\infty} \frac{(-1)^\nu (se)^{2\nu+s+\lambda+2c-k}} {2^{2\nu+s+\lambda+2c-k}\nu!(\nu+s+\lambda+2c-k)!}\\ &=&2\pi\left(\frac{se}{2}\right)^{s+\lambda}\sum_{k=0}^\infty {{k+3} \choose 3} s^{-k}\sum_{c=0}^k {k \choose c} \sum_{\nu=max(0,k-s-\lambda-2c)}^{\infty} \frac{(-1)^\nu (s^2e^2/4)^{\nu+c}}{\nu!(\nu+s+\lambda+2c-k)!}\end{split}\]

Similarly to before we would like to group by powers of the eccentricity: \(n=\nu+c\). This leads to the following constraints:

\[\begin{split}\nu>=0 & \Rightarrow & c \le n\\ \nu>=k-s-\lambda-2c & \Rightarrow & c \ge k-s-\lambda-n\\ k-s-\lambda-n \le n & \Rightarrow & k \le 2n+\lambda+s\\ k-s-\lambda-n \le k & \Rightarrow & n \ge -s - \lambda\end{split}\]

Plugging into the expression above:

\[I_{\lambda,s} = \sum_{n=\max(0,-s-\lambda)}^\infty \beta_{\lambda,s,n} (se/2)^{s+\lambda+2n}\]

with

\[\beta_{\lambda,s,n}\equiv \frac{(-1)^n}{\omega} \sum_{k=0}^{2n+\lambda+s} {{k+3} \choose 3} s^{-k}\sum_{c=\max(0,k-\lambda-s-n)}^{\min(n,k)} {k \choose c} \frac{(-1)^c}{(n-c)!(n+\lambda+s+c-k)!}\]

In terms of \(I_{\lambda,s}\):

\[p_{\pm2,s}=\exp(\mp 2i\phi_0)\left\{p_{0,s} +(1-e^2)\left[(I_{2,s}+I_{-2,s})/2-I_{0,s}\right] \mp \sqrt{1-e^2}(I_{2,s}-I_{-2,s})/2 \pm e\sqrt{1-e^2}(I_{1,s}-I_{-1,s}) \right\}\]

Verified using Methematica for \(s\neq0\).

Using:

\[\sqrt{1-e^2}=\sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2(1-2n)} e^{2n}\]

we can rewrite:

\[p_{\pm2,s}=\exp(\mp 2i\phi_0)\sum_{n=-1}^\infty \gamma^\pm_{s,n}\left(\frac{se}{2}\right)^{2n+s}\]

with:

\[\begin{split}\gamma^\pm_{s,n} &\equiv &\alpha_{s,n} + \frac{\beta_{2,s,n-1}+\beta_{-2,s,n+1}}{2} - \beta_{0,s,n}+\frac{4}{s^2}\beta_{0,s,n-1} - \frac{2}{s^2}\left(\beta_{2,s,n-2}+\beta_{-2,s,n}\right)\\ &&{}\pm \sum_{k=0}^{n+1} \frac{(2k)!}{s^{2k}(k!)^2(2k-1)} \left[\frac{1}{2}\left(\beta_{2,s,n-k-1}-\beta_{-2,s,n-k+1}\right)+ \frac{2}{s}\left(\beta_{-1,s,n-k}-\beta_{1,s,n-k-1}\right)\right]\end{split}\]

Verified by Mathematica.

Plugging in the bessel function expressions:

\[\begin{split}p_{\pm2,s}&=&\frac{\exp(\mp 2i\phi_0)}{\omega}\sum_{k=0}^\infty \left(\frac{e}{2}\right)^k \sum_{c=0}^k {k \choose c} \Bigg\{ (k+1)J_{s+2c-k}(es) + \\ &&{}+{{k+3} \choose 3} \Bigg[ -(1-e^2)J_{s+2c-k}(es) +\frac{1-e^2}{2}\big[J_{s+2+2c-k}(es)+J_{s-2+2c-k}(es)\big] \mp\\ &&\quad\quad\quad\quad\quad{} \mp\frac{\sqrt{1-e^2}}{2}\big[J_{s+2+2c-k}(es)- J_{s-2+2c-k}(es)\big] \pm e\sqrt{1-e^2}\big[J_{s+1+2c-k}(es)-J_{s-1+2c-k}(es)\big] \Bigg] \Bigg\}\end{split}\]

For s=0 we need to go back to:

\[\begin{split}2\pi p_{\pm2,0}&=&\exp\left(\mp 2i\phi_0\right) \left\{2\pi p_{0,0} + \frac{1}{\omega}\left[ (1-e^2)\int_{0}^{2\pi} \frac{\cos 2u -1} {(1-e\cos u)^4} du \mp i\sqrt{1-e^2}\int_{0}^{2\pi} \frac{\sin 2u} {(1-e\cos u)^4} du \pm i2e\sqrt{1-e^2} \int_{0}^{2\pi} \frac{\sin u} {(1-e\cos u)^4} du \right]\right\}\\ &=&\exp\left(\mp 2i\phi_0\right) \left\{2\pi p_{0,0} + \frac{2}{\omega}\left[ (1-e^2)\int_{0}^{2\pi} \frac{\cos^2 u -1} {(1-e\cos u)^4} du \pm i\sqrt{1-e^2}\int_{0}^{2\pi} \frac{\cos u}{(1-e\cos u)^4}d\cos u \mp ie\sqrt{1-e^2} \int_{0}^{2\pi} \frac{1}{(1-e\cos u)^4} d\cos u \right]\right\}\end{split}\]

\[\begin{split}\int_{0}^{2\pi}\frac{\cos^{2n} u} {(1-e\cos u)^4} du &=&\sum_{k=0}^\infty {2k+3 \choose 3} e^{2k} \int_{0}^{2\pi}\frac{\cos^{2n+2k} u} du\\ &=&2\pi\sum_{k=0}^\infty {2k+3 \choose 3} \frac{(2k+2n)!}{2^{2k+2n}[(k+n)!]^2} e^{2k}\end{split}\]

\[\begin{split}\int_{0}^{2\pi} \frac{1}{(1-e\cos u)^4} d\cos u &=&-\frac{1}{e}\int_{0}^{2\pi} \frac{1}{(1-e\cos u)^4} d(1-e\cos u)\\ &=&\left.\frac{1}{3e(1-e\cos u)^3}\right|_{0}^{2\pi}\\ &=&0\end{split}\]

\[\begin{split}\int_{0}^{2\pi} \frac{\cos u}{(1-e\cos u)^4}d\cos u &=&\frac{1}{e^2}\int_{0}^{2\pi} \frac{1-e\cos u-1}{(1-e\cos u)^4} d(1-e\cos u)\\ &=&\frac{1}{e^2}\int_{0}^{2\pi} \frac{1}{(1-e\cos u)^3} d(1-e\cos u)\\ &=&-\frac{1}{2e^2(1-e\cos u)^2}\\ &=&0\end{split}\]

So we are left with:

\[\begin{split}p_{\pm2,0}&=&\exp\left(\mp 2i\phi_0\right) \left\{p_{0,0} + \frac{2(1-e^2)}{\omega}\left\{\sum_{k=0}^\infty {2k+3 \choose 3} \frac{(2k+2)!}{2^{2k+2}[(k+1)!]^2}- \frac{2k!}{2^{2k}(k!)^2}\right\}e^{2k}\right\}\\ &=&\exp\left(\mp 2i\phi_0\right) \left\{p_{0,0} - \frac{2(1-e^2)}{\omega}\sum_{k=0}^\infty {2k+3 \choose 3} \frac{2k!}{2^{2k}(k!)^2(2k+2)}e^{2k}\right\}\\ &=&\exp\left(\mp 2i\phi_0\right) \left\{p_{0,0} - \frac{2}{\omega}\sum_{k=0}^\infty \left[ {2k+3 \choose 3} \frac{2k!}{2^{2k}(k!)^2(2k+2)} - {2k+1 \choose 3} \frac{2(k-1)!}{2^{2k-2}[(k-1)!]^2 2k} \right]e^{2k}\right\}\\ &=&\exp\left(\mp 2i\phi_0\right) \left\{p_{0,0} - \frac{2}{\omega}\sum_{k=0}^\infty \left[ \frac{(2k+3)(2k+1)!}{6\,2^{2k}(k!)^2} - \frac{4k^2(2k+1)(2k-1)!}{6\,2^2k(k!)^2} \right]e^{2k}\right\}\\ &=&\exp\left(\mp 2i\phi_0\right) \left\{p_{0,0} - \frac{2}{\omega}\sum_{k=0}^\infty \left[ \frac{(2k+3)(2k+1)!}{6\,2^{2k}(k!)^2} - \frac{2k(2k+1)!}{6\,2^2k(k!)^2} \right]e^{2k}\right\}\\ &=&\exp\left(\mp 2i\phi_0\right) \left\{p_{0,0} - \frac{2}{\omega}\sum_{k=0}^\infty \frac{(2k+1)!}{2^{2k+1}(k!)^2} e^{2k}\right\}\\ &=&0\end{split}\]

Confirmed by Mathematica.