Alternative series expansion for the elliptic integral of the second kind

Question

I found the following alternative series expansion for the complete elliptic integral of the second kind (E):

$ E(k) = \frac{(1+k')\pi}{4} \left\{ 1+\frac{1}{2^2}\left(\frac{1-k'}{1+k'}\right)^2+ \frac{1^2}{2^2\cdot 4^2} \left(\frac{1-k'}{1+k'}\right)^4 +...+ \left( \frac{(2n-3)!!}{2^n n!}\right)^2 \left(\frac{1-k'}{1+k'}\right)^{2n} +...\right\} $

where $k'=\sqrt{1-k^2}$, $\; k$ is the modulus and $k'$ is the complementary modulus.

I have been trying to prove it without success. This expansion seems to be related to the Landen transformation:

$K\left(\frac{1-k'}{1+k'}\right) = \frac{1+k'}{2}K(k)$

$ E\left(\frac{1-k'}{1+k'}\right) = \frac{1}{1+k'}\left[E(k)+k'K(k) \right] $

where K(k) is the complete elliptic integral of the first kind.

Here is my try:

From the last relations we have:

$ E(k) = (1+k')E\left(\frac{1-k'}{1+k'}\right) -k'K(k) = (1+k')E\left(\frac{1-k'}{1+k'}\right) -\frac{2k'}{1+k'}K\left(\frac{1-k'}{1+k'}\right)$

Then I used the classic expansion of $E$ and $K$:

$K(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2k^{2n} $

$E(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{1}{1-2n} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2k^{2n} $

To get:

\begin{align*} E(k) = & (1+k')\frac{\pi}{2} \sum_{n=0}^{\infty} \frac{1}{1-2n} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-k'}{1+k'}\right)^{2n} -\frac{2k'}{1+k'} \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-k'}{1+k'}\right)^{2n}\\ =& \frac{\pi(1+k')}{2} \sum_{n=0}^{\infty}\left\{ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-k'}{1+k'}\right)^{2n} \left(\frac{1}{1-2n} - \frac{2k'}{(1+k')^2}\right)\right\} \end{align*} but the path came to a dead end. I will really appreciate your help.

Answer 1

Your path is not a dead end! We have $\frac{2k'}{(1+k')^2} = \frac{1}{2}\left[1 - \left(\frac{1-k'}{1+k'}\right)^2\right]$, so your last line implies \begin{align} \frac{4 \operatorname{E}(k)}{\pi (1+k')} &= \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\left[\frac{2}{1-2n} - 1 + \left(\frac{1-k'}{1+k'}\right)^2\right] \\ &= 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2(n+1)} \\ &= 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n-2)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n} \\ &= 1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n} \left[(2n)^2 - \frac{2n+1}{2n-1} (2n-1)^2\right] \\ &= 1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n} = \sum \limits_{n=0}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\, . \end{align}