Calculation of the Pm,s Coefficients

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

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

For \(m=0\):

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

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

\[\begin{split}p_{0,s}&=&\frac{1}{\omega}\int_0^{2\pi} \left[\cos u + i \sin u\right]^s \left[\sum_{l=0}^\infty \frac{(-ise)^l\sin^l u}{l!}\right] \left[\sum_{p=0}^\infty (p+1) e^p\cos^p u\right]\\ &=&\frac{1}{\omega}\int_0^{2\pi} \left[\sum_{k=0}^s {s \choose k} i^k\cos^{s-k} u \sin^k u\right] \left[\sum_{l=0}^\infty \frac{(-ise)^l\sin^l u}{l!}\right] \left[\sum_{p=0}^\infty (p+1) e^p\cos^p u\right]\\ &=&\frac{1}{\omega}\sum_{k=0}^s \sum_{l=0}^\infty \sum_{p=0}^\infty (-1)^l i^{k+l} (p+1) {s \choose k} \frac{s^l}{l!} e^{l+p} I_{k+l,s-k+p}\\ &=&\frac{1}{\omega}\sum_{n=0}^\infty \left[\sum_{k=0}^s {s \choose k} \sum_{l=0}^n (-1)^l i^{k+l} (n-l+1) \frac{s^l}{l!} I_{k+l,s-k+n-l}\right]e^n\end{split}\]

with

\[I_{m,n} \equiv \int_0^{2\pi} \sin^m u \cos^n u\]

To find an expression for \(m=\pm2\) consider:

\[\begin{split}Q_{p,q}&\equiv& \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]} {(1-e\cos u)^4} \sin^p u \cos^q udu\\ &=& \sum_{\lambda=0}^\infty \frac{(-ise)^\lambda}{\lambda!} \int_{0}^{2\pi} \frac{\sin^{\lambda+p} u \cos^q u \left(\cos u + i\sin u\right)^s}{(1-e\cos u)^4}\end{split}\]

Using: \(1/(1-x)^4=\sum_{k=0}^\infty {{k+3} \choose 3} x^k\)

\[\begin{split}Q_{p,q}&=&\sum_{\lambda=0}^\infty \frac{(-ise)^\lambda}{\lambda!} \sum_{k=0}^\infty {k+3 \choose 3} e^k \sum_{\sigma=0}^{s} {s \choose \sigma} i^\sigma I_{\lambda+\sigma+p, s-\sigma+k+q}\\ Q_{p,q}&=&\sum_{n=0}^\infty \left[ \sum_{\lambda=0}^n \frac{(-is)^\lambda}{\lambda!} {n-\lambda+3 \choose 3} \sum_{\sigma=0}^{s} {s \choose \sigma} i^\sigma I_{\lambda+\sigma+p, s+n+q-\sigma-\lambda}\right] e^n\end{split}\]

here we show:

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

According to this page

\[\sqrt{1-e^2}=1-\frac{e^2}{2}- \sum_{n=2}^\infty\frac{(2n-3)!}{2^{2n-2} n! (n-2)!} e^{2n}\]

Also note that:

\[\begin{split}I_{m,n}&=&I_{m,n-2}-I_{m+2,n-2}=I_{m,n-2}-\frac{m+1}{n-1}I_{m,n}\\ \Rightarrow I_{m,n}&=&\frac{n-1}{n+m}I_{m,n-2}\end{split}\]

Similarly:

\[\begin{split}I_{m,n}&=&I_{m-2,n}-I_{m-2,n+2}=I_{m-2,n}-\frac{n+1}{m-1}I_{m,n}\\ \Rightarrow I_{m,n}&=&\frac{m-1}{n+m}I_{m-2,n}\end{split}\]

Further \(I_{m,1}=I_{1,n}=0\), so \(I_{m,n}\) is non-zero only if both m and n are even.

Finally, \(I_{m,n}=I_{n,m}\).