Infinite series representation of elliptic integrals $F(k,\phi)$ and $E(k,\phi).$

Here is my working. For a natural number $n,$$2^{2n-1}\left(-1\right)^n\sin^{2n}\phi=\tfrac{1}{2}\left(-1\right)^n\binom{2n}{n}+\sum\limits_{r=0}^{n-1}\binom{2n}{r}\left(-1\right)^r\cos 2\left(n-r\right)\phi.$$ According to the binomial theorem, $$\frac{1}{\sqrt{1-x^2}}=1+\tfrac{1}{2}x^2+\tfrac{3}{8}x^4+\cdots+\tfrac{\left(2n-1\right)!!}{\left(2n\right)!!}x^{2n}+\cdots$$ and upon substituting$k\sin\phi$for$x$and integrating we get $$\phi+\sum\limits_{n=1}^{\infty}\frac{\left(2n-1\right)!!k^{2n}}{\left(2n\right)!!2^{2n-1}}\left\lbrace\frac{\phi}{2}\binom{2n}{n}+\sum\limits_{r=0}^{n}\binom{2n}{r}\frac{\left(-1\right)^{n+r}\sin 2\left(n-r\right)\phi}{2\left(n-r\right)}\right\rbrace=F(k,\phi)?$$ Similarly, since$\sqrt{1-x^2}$has a series expansion with opposite sign past the first term, and the general term has$\left(2n-3\right)!!$in place of$\left(2n-1\right)!!,$$$\phi-\sum\limits_{n=1}^{\infty}\frac{\left(2n-3\right)!!k^{2n}}{\left(2n\right)!!2^{2n-1}}\left\lbrace\frac{\phi}{2}\binom{2n}{n}+\sum\limits_{r=0}^{n}\binom{2n}{r}\frac{\left(-1\right)^{n+r}\sin 2\left(n-r\right)\phi}{2\left(n-r\right)}\right\rbrace=E(k,\phi)?$$ I would like to know if somebody could confirm or correct these formulae, since I only have some tabulated series in a book to go on and these are not listed as far as I can see. I believe the series for$F(k,\tfrac{\pi}{2})$gives the following expression: $$\int^{\tfrac{\pi}{2}}_0\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}=\frac{\pi}{2}\left\lbrace1+\left(\frac{1}{2}\right)^2k^2+\left(\frac{1\times 3}{2\times 4}\right)^2k^4+\left(\frac{1\times 3\times 5}{2\times 4\times 6}\right)^2k^6+\cdots\right\rbrace,$$ and the second gives $$\int^{\tfrac{\pi}{2}}_0\sqrt{1-k^2\sin^2\phi}d\phi=\frac{\pi}{2}\left\lbrace 1-\left(\frac{1}{2}\right)^2k^2-\left(\frac{1\times 3}{2\times 4}\right)^2\frac{k^4}{3}-\left(\frac{1\times 3\times 5}{2\times 4\times 6}\right)^2\frac{k^6}{5}-\cdots\right\rbrace$$ which I think is correct, but I would like somebody to confirm the general formula, ie. the series representations for the elliptic integrals if possible. I cannot find any explicit representations of these integrals as series. I'm aware this is a pretty simple question but I just need a confirmation of either being right or wrong, no need for fancy answers even though they're welcome. • Did you check wiki page ? The same result you got for K(k) series expansion en.wikipedia.org/wiki/Elliptic_integral – Mathlover Apr 26 '14 at 12:24 • I have seen the specific result for$F(k,\pi/2)$but it is the general formulae which I am not sure about. I'm just hoping to check that there is not some mistake. – Hobbyist Apr 26 '14 at 12:31 1 Answer Assuming that your last line is (you have a typo in the integrand), I just confirm that your development is perfectly correct $$F(k,\pi/2)=\int^{\pi/2}_0\frac{d\phi}{\sqrt{1-k^2\sin^2 \phi}}=\frac{K\left(\frac{k^2}{k^2-1}\right)}{\sqrt{1-k^2}}$$ for which the Taylor expansion built at$k=0\$ effectively leads to $$F(k,\pi/2)=\frac{\pi}{2}\left\lbrace1+\left(\frac{1}{2}\right)^2k^2+\left(\frac{1\times 3}{2\times 4}\right)^2k^4+\left(\frac{1\times 3\times 5}{2\times 4\times 6}\right)^2k^6+\cdots\right\rbrace$$

Nice work and well done !