I am attempting to solve an impossibly difficult (to me) integral, if you can offer any advice or help in any way then please do; your assistance is greatly appreciated.
The integral(s) are;
$$I(\ell',\ell) = \int_0^{2\pi} \int_0^{\pi}\,{\rm d}\phi\,{\rm d}\theta \ \ Y_{\ell',m}^{*}(\theta,\phi) \, Y_{\ell,m}(\theta,\phi)\ \ \frac{1}{\sqrt{1-(1-\frac{\gamma}{\beta})\cos(\theta)^2}} $$
Where $ Y_{\ell,m}(\theta,\phi)$ is the spherical harmonic for $\ell,m$, and $^*$ denotes the complex conjugate.
Even confining myself to the $(\ell,\ell',m) = (0,0,0)$ case is not obvious to me; Since $Y_{0,0}^{*}(\theta,\phi) = Y_{0,0}(\theta,\phi) = \frac{1}{2\sqrt{\pi}}$ we have
$$
I(0,0) = \int_0^{2\pi} \int_0^{\pi}\,{\rm d}\phi \,{\rm d}\theta \,\frac{1}{4\pi} \frac{1}{\sqrt{1-(1-\frac{\gamma}{\beta})\cos(\theta)^2}} \\
$$
$$
= \int_0^{\pi}\, {\rm d}\theta \, \frac{2\pi}{4\pi} \frac{1}{\sqrt{1-(1-\frac{\gamma}{\beta})\cos(\theta)^2}} \\
$$
$$
= \int_0^{\pi}\,{\rm d}\theta \, \frac{1}{2} \frac{1}{\sqrt{1-(1-\frac{\gamma}{\beta})\cos(\theta)^2}}
$$
I've never been particularly adept at spotting the "trick" substitution. As far as I can see, any substitution involving $\cos(\theta)$ will result in terms dependent upon $\theta$ which then invalidates the point of the substitution in the first place.
For the curious, the function $I(\ell',\ell)$ is taken from equation 2.14 of the paper by R. A. Faulkner 1969.
****EDIT** based upon the suggestions of @tired and @winther about this being similar to an elliptical function.**
By making the substitution $\cos(\theta)^2 = 1- \sin(\theta)^2$;
$$\sqrt{1-(1-\frac{\gamma}{\beta})\cos(\theta)^2} = \sqrt{1-(1-\frac{\gamma}{\beta})(1- \sin(\theta)^2)}$$
$$=\sqrt{\frac{\gamma}{\beta} + (1-\frac{\gamma}{\beta})\sin(\theta)^2}$$
$$=(\frac{\gamma}{\beta})^{\frac{1}{2}} \sqrt{1 + (\frac{\beta}{\gamma}-1)\sin(\theta)^2}$$
$$=(\frac{\gamma}{\beta})^{\frac{1}{2}} \sqrt{1 – (1-\frac{\beta}{\gamma})\sin(\theta)^2}$$
which is the same form as the "complete elliptic integral of the first kind" for the $\ell',\ell,m=0,0,0$ case.
So that we get;
$$
I(0,0)=\frac{1}{2(\frac{\gamma}{\beta})^{\frac{1}{2}}} \int_0^{\pi}\,{\rm d}\theta \, \frac{1}{\sqrt{1 – (1-\frac{\beta}{\gamma})\sin(\theta)^2}}
$$
With the complete elliptic integral of the first kind being;
$$K(k)=\int_0^{\frac{\pi}{2}} \frac{{\rm d}\theta}{\sqrt{[1-k^{2}\sin(\theta)^2]}} $$
So, if I'm not mistaken (This is where you correct me), as between $0$ and $\pi$ sin is symmetrical about $\frac{\pi}{2}$, giving a factor of 2, then;
$$ I(0,0) = \frac{1}{2(\frac{\gamma}{\beta})^{\frac{1}{2}}} K( (1-\frac{\beta}{\gamma}) )$$ ?
And then for the general case, I am left with;
$$I(\ell',\ell) =\frac{1}{(\frac{\gamma}{\beta})^{\frac{1}{2}}} \int_0^{2\pi} \int_0^{\pi}\,{\rm d}\phi\,{\rm d}\theta \ \frac{Y_{\ell',m}^{*}(\theta,\phi) \, Y_{\ell,m}(\theta,\phi)}{\sqrt{1-(1-\frac{\gamma}{\beta})\sin(\theta)^2}} $$
Best Answer
Let's do one of the non-elementary integrals (we assume $\alpha^2\equiv1-\gamma/\beta>0$):
Take $(2,0,m)$. then the integral takes the form (due to complex conjugation the integral corresponding to $m$ will be always trivial)
$$ I(2,0,m)=C\int_0^\pi d\theta \frac{3\cos(\theta)^2-1}{\sqrt{1-\alpha^2\cos(\theta)^2}} $$
Here $C$ is some constant.
Now employing a substitution $\theta\rightarrow \Theta-\pi/2$ and usinng the symmetry of the integrand this boils down to
$$ I(2,0,m)=2C\int_0^{\pi/2} d\Theta \frac{3\sin(\Theta)^2-1}{\sqrt{1-\alpha^2\sin(\Theta)^2}}=2C\left(\int_0^{\pi/2}d\Theta \frac{3\sin(\Theta)^2}{\sqrt{1-\alpha^2\sin(\Theta)^2}}-\int_0^{\pi/2}d\Theta \frac{1}{\sqrt{1-\alpha^2\sin(\Theta)^2}}\right)\\=2C\left(\frac{-3}{\alpha^2}\int_0^{\pi/2}d\Theta \frac{1-\alpha^2\sin(\Theta)^2-1}{\sqrt{1-\alpha^2\sin(\Theta)^2}}-\int_0^{\pi/2}d\Theta \frac{1}{\sqrt{1-\alpha^2\sin(\Theta)^2}}\right)\\ =2C\left(\frac{-3}{\alpha^2}\int_0^{\pi/2}d\Theta \sqrt{1-\alpha^2\sin(\Theta)^2}+\left(\frac{3}{\alpha^2}-1\right)\int_0^{\pi/2}d\Theta \frac{1}{\sqrt{1-\alpha^2\sin(\Theta)^2}}\right) $$
Employing the definiton of the complete elliptic integrals of the first and second kind we can rewrite this as
$$ I(2,0,m)=-\frac{6C}{\alpha^2}E(\alpha)+2C\left(\frac{3}{\alpha^2}-1\right)K(\alpha) $$