Ellipsoidal harmonics - A Series expansion for Lame functions of the second kind

Question

$\underline{Intro \;to \;skip}$

In the theory of ellipsoidal harmonics, Lame functions of the second kind $F_n$ arise as the second linearly independent solution (the first being Lame functions of the first kind $E_n$) of Laplace's equation in ellipsoidal coordinates which is the so called Lame's equation. Lame functions of the second kind contain the improper elliptic integral $I_n(\rho)$ and for this reason they are obtainable through numerical approximations e.g. Gauss-Legrendre quadrature. For anyone interested in ellipsoidal harmonics the main bibliographical references (as fas as I know) are Ellipsoidal Harmonics by Dassios, The Theory of Spherical and Ellipsoidal Harmonics by Hobson, An Elementary Treatise on Fourier's Series by Byerly.

$\underline{Question}$

My goal is to find an expansion in powers of 1/ρ (and its first 2 or 3 terms) of the quantity \begin{equation}\label{eq:1} F(\rho,\mu,\nu)=(2n+1)E_n(\rho)E_n(\mu)E_n(\nu)I_n(\rho),\quad \rho \ge h_2\end{equation} where \begin{equation} I_n(\rho)=\int_\rho^{+\infty}\frac{dt}{(E_n(t))^2\sqrt{t^2-h_2^2}\sqrt{t^2-h_3^2}} \end{equation} $\rho,\mu,\nu$ are the ellipsoidal coordinates and \begin{equation} E_n(t)=t^{n-2r}\sum_{i=0}^ra_it^{2i}, \end{equation} where \begin{align} r&=n/2, \quad n \quad \text{even},\\ r&=(n-1)/2, \quad n \quad \text{odd}. \end{align}

$\underline{What\; have\;I\; tried}$

A first step to solution is to set $t\rightarrow \frac{h_2}{t}$ and this leads to \begin{equation} I_n(\rho)=\int_0^\frac{h_2}{\rho}\frac{t^{2n}/h_2^{2n}}{(E_n(t))^2\sqrt{1-t^2}\sqrt{1-\frac{h_3^2}{h_2^2}t^2}}dt. \end{equation} Using binomial theorem \begin{align} (1-t^2)^{-1/2}&=\sum_{n=0}^{+\infty}(-1)^n\binom{-\frac{1}{2}}{n}t^{2n},\\ (1-\frac{h_3^2}{h_2^2}t^2)^{-1/2}&=\sum_{n=0}^{+\infty}(-1)^n\binom{-\frac{1}{2}}{n}\frac{h_3^{2n}}{h_2^{2n}}t^{2n}, \end{align} and multiplying gives the series expansion \begin{equation}\sum_{n=0}^{+\infty}\sum_{k=0}^n(-1)^n\binom{-\frac{1}{2}}{n-k}\binom{-\frac{1}{2}}{k}\frac{h_3^{2k}}{h_2^{2k+1}}t^{2n}\end{equation} so: \begin{equation}I_n(\rho)=\int_0^{h_2/\rho}\left[\frac{t^{2n}/h_2^{2n}}{(E_n(t))^2 }\sum_{n=0}^{+\infty}\sum_{k=0}^n(-1)^n\binom{-\frac{1}{2}}{n-k}\binom{-\frac{1}{2}}{k}\frac{h_3^{2k}}{h_2^{2k+1}}t^{2n}\right] dt .\end{equation}

$\underline{The\;problem}$

I don t know how to take it from here. The basic problem is that by trying to express everything as a sum in the quantity of F which is the brute force/obvious way, it gets so complicated to perform multiplication between these sums that I don't know if I could find my 2-3 first terms by the resulting series expansion.

Hope you like the subject and offer some help on the matter. Thank you very much :)

Answer 1

just to follow up on Michael Renardy's suggestion: let's take $n$ even $>2$, then the integrand in $I_n(t)$ has the small-$t$ expansion [using $E_n(t)=a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)$, with $n$-dependent coefficients $a_p$]:

$$K(t)=\frac{(t/h_2)^{2n}}{E_n(t)^2\sqrt{1-t^2}\sqrt{1-h_3^2(t/h_2)^2}}=c_0 (t/h_2)^{2n}+c_1 (t/h_2)^{2n+2}+c_2 (t/h_2)^{2n+4} + {\cal O}(t^{2n+6})$$

with coefficients

$$c_0=\frac{1}{a_0^2},\;\;c_1=\frac{a_0 (h_2^2+h_3^2)-4 a_1 h_2^2}{2 a_0^3},$$

$$c_2=\frac{ h_2^4\left(3 a_0^2-8 a_0 a_1-16a_0 a_2+24 a_1^2\right)+3 a_0^2 h_3^4+2 a_0 h_2^2 h_3^2 (a_0-4 a_1)}{8 a_0^4}$$

integration $I_n(\rho)=\int_0^{h_2/\rho}K(t)dt$ gives you the desired first three terms of the expansion for $n$ even,

$$I_n(\rho)=\frac{c_0 h_2 }{2n+1}(1/\rho)^{2n+1}+\frac{c_1 h_2}{2n+3}(1/\rho)^{2n+3}+\frac{c_2 h_2}{2n+5}(1/\rho)^{2n+5}+{\cal O}(1/\rho)^{2n+7}$$

for $n$ odd $>3$ we have instead $E_n(t)=t\times[a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)]$ so the expansion becomes

$$I_n(\rho)=\frac{c_0/ h_2 }{2n-1}(1/\rho)^{2n-1}+\frac{c_1/ h_2}{2n+1}(1/\rho)^{2n+1}+\frac{c_2/ h_2}{2n+3}(1/\rho)^{2n+3}+{\cal O}(1/\rho)^{2n+5}$$