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Calculation of the Pm,s Coefficients {#InclinationEccentricity_pms1}
====================================

We need only \f$m=0\f$ and \f$m=\pm2\f$.
Clearly:
\f{eqnarray*}{
    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
\f}
For \f$m=0\f$:
\f{eqnarray*}{
    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
\f}
From \f$1/(1-x)^2=\sum_{k=0}^\infty (k+1)x^k\f$:
\f[
    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
\f]
Which for \f$s=0\f$, using
 \f$\int_0^{2\pi} \cos^{2k} u du = \frac{2\pi (2k)!}{2^{2k}(k!)^2}\f$ gives:
\f[
    p_{0,0}=\frac{1}{\omega}
        \sum_{k=0}^\infty \frac{(2k+1)!}{2^{2k}(k!)^2}e^k
\f]
And for \f$s \neq 0\f$:
\f{eqnarray*}{
    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
\f}
If we change variable \f$u=u'-\pi/2\Rightarrow \sin u = -\cos u'\f$:
\f{eqnarray*}{
    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)!}\\
\f}
If we now change indices to
 \f$n=\lambda+c\Rightarrow \lambda=n-c,\quad\lambda+s+2c-k=n+s+c-k\f$. The
lower limit on \f$\lambda\f$ gives:
 \f$k-s-2c\leq n-c\Rightarrow c\geq k-n-s\f$. Finally, in order for the range
of \f$c\f$ to not be empty: \f$k-n-s\leq n\Rightarrow k\leq 2n+s\f$. With all
these we can write:
\f[
    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.
\f]
with
\f[
    \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.
\f]
Verified using Mathematica.

For \f$m=\pm2\f$ we need:
\f{eqnarray*}{
    \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}
\f}
Plugging into the expression for \f$p_{\pm2,s}\f$:
\f{eqnarray*}{
    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
\f}
Thus we need to evaluate 5 different integrals, the first of which was
already done while calculating \f$p_{0,s}\f$:
\f{eqnarray*}{
    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\\
\f}
To solve them we will use
 \f$1/(1-x)^4=\sum_{k=0}^\infty {{k+3} \choose 3} x^k\f$ and we will directly
calculate the following general integral:
\f{eqnarray*}{
    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)!}
\f}
Similarly to before we would like to group by powers of the eccentricity:
\f$n=\nu+c\f$. This leads to the following constraints:
\f{eqnarray*}{
    \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
\f}
Plugging into the expression above:
\f[
    I_{\lambda,s} = \sum_{n=\max(0,-s-\lambda)}^\infty \beta_{\lambda,s,n}
        (se/2)^{s+\lambda+2n}
\f]

with
\f[
    \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)!}
\f]

In terms of \f$I_{\lambda,s}\f$:
\f[
    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\}
\f]
Verified using Methematica for \f$s\neq0\f$.

Using:
\f[
    \sqrt{1-e^2}=\sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2(1-2n)} e^{2n}
\f]
we can rewrite:
\f[
    p_{\pm2,s}=\exp(\mp 2i\phi_0)\sum_{n=-1}^\infty
        \gamma^\pm_{s,n}\left(\frac{se}{2}\right)^{2n+s}
\f]
with:
\f{eqnarray*}{
    \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]
\f}
Verified by Mathematica.

Plugging in the bessel function expressions:
\f{eqnarray*}{
    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\}
\f}
For s=0 we need to go back to:
\f{eqnarray*}{
    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\}
\f}
\f{eqnarray*}{
    \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}
\f}
\f{eqnarray*}{
    \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
\f}
\f{eqnarray*}{
    \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
\f}
So we are left with:
\f{eqnarray*}{
    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
\f}
Confirmed by Mathematica.