How can i prove that the elliptic intergal is equal to this double sum? $\Pi(n|m)=\int _0^{\frac{\pi}{2}}\frac{1}{(1-n\sin^2t)\sqrt{1-m\sin^2t}}dt=..$ How can i show that those are the same?
$\Pi \left(n|m\right)=\int _0^{\frac{\pi }{2}}\frac{1}{\left(1-n\sin ^2t\right)\sqrt{1-m\sin ^2t}}dt=\frac{\pi }{2}\sum _{k=0}^{\infty }\sum _{j=0}^k\frac{\left(2k\right)!\left(2j\right)!m^jn^{k-j}}{4^k4^jk!^2j!^2},\:\left|m\right|<1\:\wedge \:\left|n\right|<1\:\:$
What i have done so far:
$\frac{\pi}{2}\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\binom{2k}{k}\frac{n^{k-j}}{4^k}\frac{m^j}{4^j}\binom{2j}{j}=S$
$=\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\binom{2k}{k}\frac{n^{k-j}}{4^k}\frac{m^j}{4^j}\frac{\pi}{2}\binom{2j}{j}=S$
By the beta function:
$\frac{\pi}{2(4^j)}\binom{2j}{j}=\int_{0}^{\frac{\pi}{2}}\sin^{2j}xdx$
$\therefore S=\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\frac{n^{k-j}}{4^k}\int_{0}^{\pi/2}(m\sin^2x)^jdx$
|m|<1 we can interchange order summation and integration
$S=\sum _{k=0}^{\infty }\binom{2k}{k}\frac{n^k}{4^k}\int _0^{\pi /2}\underbrace{\sum _{j=0}^{\infty }\:\left(\frac{m}{n}\right)^j\sin ^2xdx}_{\text{for this to converge m<n}}$
$\therefore \:S=\sum \:_{k=0}^{\infty \:}\binom{2k}{k}\frac{n^k}{4^k}\int _0^{\pi \:/2}\frac{1-\frac{m}{n}\sin ^2x\left(\frac{m^k}{n^k}\sin ^{2k}x\right)dx}{1-\frac{m}{n}\sin ^2x}$
So that i have tried so far is substituting the binomial coefficient with the integral of $\sin^2j$ then interchange sumation and integration. but i end ut with a wierd integral that would diverge at m=n. so where do i go from this? can anyone show me how i can show that they are the same?
Found it here:
http://functions.wolfram.com/EllipticIntegrals/EllipticK/introductions/CompleteEllipticIntegrals/05/
 A: $$\begin{align}
\frac{1}{\left(1-n\sin ^2t\right)\sqrt{1-m\sin ^2t}}&=
\sum_{i=0}^\infty\binom{-1}i (-n)^i\sin^{2i}t \sum_{j=0}^\infty\binom{-1/2}j (-m)^j\sin^{2j}t\\
&=\sum_{i=0}^\infty n^i\sin^{2i}t \sum_{j=0}^\infty \frac1{4^j}\binom{2j}j m^j\sin^{2j}t\\
&=\sum_{k=0}^\infty\sum_{j=0}^k\frac1{4^j}\binom{2j}j m^j n^{k-j}\sin^{2k} t.
\end{align}
$$
Integrating now both sides from $0$ to $\frac\pi2$ over $t$ we obtain the desired identity:
$$\begin{align}
\int_0^{\pi/2}\frac{dt}{\left(1-n\sin ^2t\right)\sqrt{1-m\sin ^2t}}
&=\int_0^{\pi/2}\left[\sum_{k=0}^\infty\sum_{j=0}^k\frac1{4^j}\binom{2j}j m^j n^{k-j}\sin^{2k} t\right]dt\\
&=\sum_{k=0}^\infty\sum_{j=0}^k\frac1{4^j}\binom{2j}j m^j n^{k-j}\int_0^{\pi/2}\sin^{2k} t\, dt\\
&=\frac\pi2\sum_{k=0}^\infty\sum_{j=0}^k\frac1{4^j}\frac1{4^k}\binom{2j}j \binom{2k}k m^j n^{k-j},
\end{align}
$$
where the mentioned in the question value of the integral:
$$\int_0^{\pi/2}\sin^{2k}t\,dt=\frac\pi2\frac1{4^k}\binom{2k}k$$ was used to obtain the last line.
