# Fourier series of elliptic F integral

Question. How to expand $$\mathbf{F}(\theta\mid k)=\int_0^\theta\frac{dt}{\sqrt{1-k^2\sin^2t}} \text{ (where |k|<1 and |\theta|<\pi)}$$as a Fourier series w.r.t. $$\theta$$?

I noticed that the question boils down to finding the Fourier expansion of $$f(\theta)=\frac1{\sqrt{1-k^2\sin^2\theta}}$$ as we can change the order of integration and differentiation. We can see clearly that $$f$$ is even, so $$\int_{-\pi}^\pi f(\theta)\sin n\theta d\theta=0.$$ Also, $$\int_{-\pi}^{\pi}f(\theta)\cos(2n+1)\theta d\theta=2\int_0^\pi f(\theta)\cos(2n+1)\theta d\theta\\=2\int_0^\pi f(\theta)\cos((2n+1)(\pi-\theta))d\theta=0$$ So the only hard part left is $$\int_0^{\pi/2}\frac{\cos 2n\theta}{\sqrt{1-k^2\sin^2\theta}}d\theta.$$ For $$n=0$$ I can clearly see it's $$\mathbf{K}(k)$$, the elliptic K integral.
Clearly there is a polynomial $$P_n(x)$$ s.t. $$\cos 2nx=P_n(\sin^2x)$$ with $$\deg P_n=n$$, but note that this does not make the question boil down to evaluating $$\displaystyle\int_0^{\pi/2}\frac{\sin^{2n}(t)}{\sqrt{1-k^2\sin^2 t}}dt$$ because after evaluating this elliptic-like integral, which I ensure that it evaluates to a form of $$a\mathbf{K}(k)+b\mathbf{E}(k)$$ by partial fraction expansion, where $$a,b\in\mathbb Q$$, there is a hard finite summation left involving the coefficients of $$P$$.
When $$n=1$$, $$a_n$$ equals $$2\mathbf{E}/k^2+(1-2/k^2)\mathbf{K}$$.
I don't have any further thoughts.
If we cannot find the general term with $$k$$ varying, can we at least find the coefficients when $$k^2=-1$$?

$$\int_0^\theta\dfrac{dt}{\sqrt{1-k^2\sin^2t}}$$
$$=\int_0^\theta\sum\limits_{n=0}^\infty\dfrac{(2n)!k^{2n}\sin^{2n}t}{4^n(n!)^2}~dt$$
$$=\int_0^\theta\sum\limits_{n=0}^\infty\dfrac{(2n)!k^{2n}C_n^{2n}}{16^n(n!)^2}~dt+\int_0^\theta\sum\limits_{n=1}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^m(2n)!k^{2n}C_{n+m}^{2n}\cos2mt}{2^{4n-1}(n!)^2}~dt$$
$$=\left[\sum\limits_{n=0}^\infty\dfrac{((2n)!)^2k^{2n}t}{16^n(n!)^4}\right]_0^\theta+\left[\sum\limits_{n=1}^\infty\sum\limits_{m=1}^n\dfrac{(-1)^m((2n)!)^2k^{2n-1}\sin2mt}{16^n(n!)^2(n+m)!(n-m)!}\right]_0^\theta$$
$$=\sum\limits_{n=0}^\infty\dfrac{\left(\tfrac{1}{2}\right)_n\left(\tfrac{1}{2}\right)_nk^{2n}\theta}{(n!)^2}+\sum\limits_{m=1}^\infty\sum\limits_{n=m}^\infty\dfrac{(-1)^m\left(\tfrac{1}{2}\right)_n\left(\tfrac{1}{2}\right)_nk^{2n-1}\sin2m\theta}{(n+m)!(n-m)!}$$
$$=\sum\limits_{n=0}^\infty\dfrac{\left(\tfrac{1}{2}\right)_n\left(\tfrac{1}{2}\right)_nk^{2n}\theta}{(n!)^2}+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^m\left(\tfrac{1}{2}\right)_{m+n}\left(\tfrac{1}{2}\right)_{m+n}k^{2m+2n-1}\sin2m\theta}{(2m+n)!n!}$$