series expansion of elliptic integral of the third kind

Question

The elliptic integral of the third kind is defined as follows: $$\Pi \left(n,m\right)=\int _0^{\frac{\pi }{2}}\:\frac{dx}{\left(1-n\left(\sin \left(x\right)\right)^2\right)\sqrt{1-m\left(\sin \:\:\:\left(x\right)\right)^2}}$$ how can you prove that: $$\Pi \left(n,m\right)=\frac{\pi }{2}\sum _{k=0}^{\infty }\sum _{j=0}^{k }\frac{m^jn^{k-j}}{4^k4^j}\begin{pmatrix}2k\\ \:k\:\end{pmatrix}\begin{pmatrix}2j\\ j\end{pmatrix}$$ where: $$\left|m\right|\:<1\:\ $$ $$\:\left|n\right|<1$$ Update:

By the beta function: $$\frac{\pi }{2}\frac{1}{4^j}\begin{pmatrix}2j\\ j\end{pmatrix}=\int _0^{\frac{\pi }{2}}\:\left(\sin \left(x\right)\right)^{2j}dx$$ $$\frac{\pi }{2}\sum _{j=0}^{k }\frac{m^jn^{k-j}}{4^k4^j}\begin{pmatrix}2j\\ j\end{pmatrix}=\int _0^{\frac{\pi }{2}}\:\sum _{j=0}^k\:\frac{m^jn^{k-j}\left(\sin \left(x\right)\right)^{2j}}{4^k}dx=\frac{n^k}{4^k}\int _0^{\frac{\pi }{2}}\:\frac{1-\left(\frac{m}{n}\right)^{k+1}\left(\sin \:\left(x\right)\right)^{2k+2}}{1-\left(\frac{m}{n}\right)\left(\sin \:\:\left(x\right)\right)^2}\:dx$$ substituting that back to our sum we get that it equals: $$\int _0^{\frac{\pi }{2}}\:\frac{\sum _{k=0}^{\infty }\begin{pmatrix}2k\\ k\end{pmatrix}\frac{n^k}{4^k}\left(1-\left(\frac{m}{n}\right)^{k+1}\left(\sin \:\:\left(x\right)\right)^{2k+2}\right)\:}{1-\left(\frac{m}{n}\right)\left(\sin \:\:\left(x\right)\right)^2}\:dx=$$ $$\int _0^{\frac{\pi }{2}}\:\frac{1}{\left(1-\left(\frac{m}{n}\right)\left(\sin \left(x\right)\right)^2\right)\sqrt{1-n}}\:-\frac{\left(\frac{m}{n}\right)\left(\sin \left(x\right)\right)^2}{\left(1-\left(\frac{m}{n}\right)\left(\sin \left(x\right)\right)^2\right)\sqrt{1-\left(m\right)\left(\sin \left(x\right)\right)^2}}dx$$ How is the final integral equivalent to the elliptic integral of the third kind

Answer 1

Use the change of variable $y = \sin(x)$ then expand the function of $y$ into a product of two power series, group the terms by power of $y$, use $\binom{2k}{k} = (-1)^k 4^k \binom{-1/2}{k} $, and integrate termwise

Answer 2

It can be easily derived that $$\frac{1}{\left(1-n\left(\sin \left(x\right)\right)^2\right)\sqrt{1-m\left(\sin \left(x\right)\right)^2}}=\sum _{i=0}^{\infty }\:\sum _{k=0}^{\infty }\:\begin{pmatrix}2k\\ k\end{pmatrix}\frac{n^im^k}{4^k}\left(\sin \:\left(x\right)\right)^{2\left(i+k\right)}$$ for $$i+k=j$$ where $$0 \le k\le j$$ then $$i=j-k$$ Thus we can rewrite our sum by grouping it by j to get: $$\sum _{j=0}^{\infty }\:\sum _{k=0}^j\:\begin{pmatrix}2k\\ k\end{pmatrix}\frac{n^{j-k}m^k}{4^k}\left(\sin \:\left(x\right)\right)^{2j}\:$$ we can then integrate and plug integral to get the desired sum $$\int _0^{\frac{\pi }{2}}\:\left(\sin \:\:\left(x\right)\right)^{2j}dx=\frac{\pi }{2}\begin{pmatrix}2j\\ j\end{pmatrix}\frac{1}{4^j} $$