Specific values of Elliptic integral of the second kind $E$

Question

As far as my search on the internet, I couldn't seem to find values of the Elliptic integral of the second kind. My current goal is to compute $E(\sqrt{2}-1)$

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Motivation

Let $\displaystyle d(x)=\sum_{n=0}^{\infty}\binom{2n}{n}^3 \frac{x^{2n}}{4^{3n}}$. Using transformation formulas of hypergeometric series we can show that $$d(2kk')=\left(\frac{2K}{\pi}\right)^2$$ Now if we differentiate both sides and use the fact that $\frac{dK}{dk}=\frac{E}{kk'^2}-\frac{K}{k}$, we get

$$\frac{d'(2kk')}{k'}(2k'^2-2k^2)=\frac{8KE}{\pi^2 k k'^2}-\frac{2}{k}d(2kk')$$

Which if written in term of infinite sums, we get

$$\frac{4KE}{\pi^2}=\sum_{n=0}^{\infty}\binom{2n}{n}^3\frac{(2kk')^{2n}}{4^{3n}}\{(1-2k^2)n+k'^2\}$$

These looks much like $\pi$ formulas given by Ramanujan. However, I was unable to obtain values because I don't know particular values of $E$ (The function diverges at $k=\frac{1}{\sqrt{2}}$) . I think there was a MSE post that also uses this method to obtain $\pi$ formulas, but I couldn't find the post. A similar procedure give a formula in terms of $\sum_{n=0}^{\infty}\frac{(4n)!}{n!^4}\frac{x^{2n}}{4^{4n}}$

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I found this twitter post which states that $$1-24\sum_{n=1}^{\infty}\frac{n}{e^{2\pi n \sqrt{2}}-1}=\frac{3}{\sqrt{2}\pi}+\frac{\Gamma^2(\frac{1}{8})\Gamma^2(\frac{3}{8})}{32 \pi^3}$$ This is a particular case of the Ramanujan $P(q^2)$ function, $P(e^{-2\sqrt{2}\pi})$. It is known that $$\left(\frac{2K}{\pi}\right)^2\text{ns}^2(u,k)= \left(\frac{2K}{\pi}\right)^2-\frac{4KE}{\pi^2}+\text{cosec} ^2\,z-8\sum_{n\geq 1}\frac{nq^{2n}}{1-q^{2n}}\cos 2nz$$

Where $z=\frac{\pi u} {2K}$. If we take the limit $z\to 0$ we get

$$P(q^2)=\frac{12 KE}{\pi^2}-(1+k'^2)\left(\frac{2K}{\pi}\right)^2$$

Obviously we want the case $q=e^{-\sqrt{2}\pi}$ such that $k=\sqrt{2}-1$. However I don't know how to compute $E(\sqrt{2}-1)$, and therefore I'm stuck.

The Eisenstein series $E_2(z)=P(q)$ where $q=e^{2\pi i z}$ has it's functional formula

$$E_2\left(\frac{az+b}{cz+d}\right)=(cz+d)^2E_2(z)+\frac{6c(cz+d)}{\pi i}$$ which I could possibly utilize.

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Possible approaches: $E$ hasn't been studied much in the field of elliptic integrals. The only properties I can recall is that $E$ has its Landen transformation,

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Edit 10/10/2024

The Ramanujan $P$ function is defined as $$P(q)=1-24\sum_{n=1}^{\infty}\frac{nq^n}{1-q^n}$$

Let $n\frac{K'}{K}=\frac{L'}{L}$. Ramanujan proved that $$nP(q^{2n})-P(q^2)=\frac{K L}{\pi^2}A(k) \tag{1}$$ For some algebraic function $A(k)$. Let $k=l'$ to obtain $$nP(e^{-2\pi \sqrt{n}})-P(e^{-2\pi/\sqrt{n}})=\frac{K^2}{\pi^2}A \tag{2}$$ with another formula $$nP(e^{-2\pi \sqrt{n}})+P(e^{-2\pi/\sqrt{n}})=\frac{6\sqrt{n}}{\pi}\tag{3}$$ gives

$$P(e^{2\pi \sqrt{n}})=\frac{K^2}{\pi^2}A+\frac{3}{\pi\sqrt{n}}$$ It is now obvious that the aforementioned formula is the case where $n=2$. Now I just have to find what $A$ is (It should be $4-2\sqrt{2}$). I have a trouble following the proof to show that $(1)$ is true, hope I could solve it soon.

Answer 1

Let us start with the following identity $$2P(q^4)-P(q^2)=\left(\frac{2K}{\pi}\right)^2\frac{1+k'^2}{2}\tag{1}$$ with $q=e^{-\pi/\sqrt{2}}$ and $k'=\sqrt{2}-1$ to get $$2P(q^4)-P(q^2)=\left(\frac{2K}{\pi}\right)^2(2-\sqrt{2}) $$ Let us further note that $K'/K=1/\sqrt {2}$ and hence $$2P(q^4)-P(q^2)=\left(\frac{2K'}{\pi}\right)^2(4-2\sqrt{2})$$ And let us combine this with another identity $$2P(q^4)+P(q^2)=\frac{6\sqrt{2}}{\pi}$$ to get $$P(q^4)=\frac{3}{\pi\sqrt{2}}+\left(\frac{2K'}{\pi}\right)^2\left(1-\frac{1}{\sqrt{2}}\right)$$ Rearranging the symbols such that $q=e^{-\pi\sqrt{2}}$ the above equation can be written as $$P(q^2)=\frac{3}{\pi\sqrt{2}}+\left(\frac{2K}{\pi}\right)^2\left(1-\frac{1}{\sqrt{2}}\right)$$ And we can compare this with $$P(q^2)= \left(\frac{2K}{\pi}\right)^2\left(\frac{3E}{K}+k^2-2\right)$$ with $k=\sqrt{2}-1$ to get the value $E$ in terms of $K$. The value of $K$ is evaluated in this thread.

Answer 2

This is a different approach to obtain the link between $K(k_{2})$ and $E(k_{2})$ when $k_{2}=\sqrt{2}-1$, however is not as general as Ramamujan's technique and only works for a few particular cases. We need the following:

1. Legendre's relation: $$K(k')E(k)+K(k)E(k')-K(k)K(k')=\frac{\pi}{2}\tag{1},$$ where $k'=\sqrt{1-k^2}$ is the complementary modulus.

2. Landen's (descending) transformations: $$K(k)(1+k)=K(\frac{2\sqrt{k}}{1+k}), \hspace{2cm}\frac{2E(k)}{1+k}+K(k)(k-1)=E(\frac{2\sqrt{k}}{1+k})\tag{2}.$$

3. And the constraint (being singular moduli): $$\frac{K(k_{n}')}{K(k_{n})}=\sqrt{n},\tag{3}$$ where n is a natural number.

With this information we can rewrite $(1)$ as: $$\sqrt{n}\left(K^2(k)-\frac{K(k)E(k)}{1+k}\right)=-\frac{\pi}{2(k+1)}+K(k)E(\frac{1-k}{1+k})\tag{4}.$$ Then if $k_{2}=\sqrt{2}-1$ we have $\frac{1-k_{2}}{1+k_{2}}=k_{2}$ and $n=2$. Replacing this values in $(4)$ gives: \begin{align*}\sqrt{2}\left(K(k_{2})-\frac{E(k_{2})}{1+k_{2}} \right)=-\frac{\pi}{2(k_{2}+1)K(k_{2})}+E(k_{2})\implies E(k_{2})=\frac{\sqrt{2}\pi}{8K(k_{2})}+\frac{\sqrt{2}K(k_{2})}{2}\tag{5}. \end{align*}

Or which is the same:

$$8E(k_{2})K(k_{2})-4\sqrt{2}K^2(k_{2})=\sqrt{2}\pi\tag{6}.$$

This links $K(k_{2})$ and $E(k_{2})$ without knowing its explicit values. We have derived Bauer's series with this approach How to Evaluate $\sum_{n=0}^{\infty}\frac{(-1)^n(4n+1)(2n)!^3}{2^{6n}n!^6}$.

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Also see: https://mathworld.wolfram.com/EllipticAlphaFunction.html The elliptic alpha function gives the link between $E(k_{n})$ and $K(k_{n})$ without knowing the explicit values of $K(k_{n})$.

In this page some $K(k_{n})$ are given: https://mathworld.wolfram.com/EllipticIntegralSingularValue.html

So there is no explicit evaluation for $E(k_{n})$ in Wolfram but the relation between $K(k_{n})$ and $E(k_{n})$ is listed via this function for some values of $n$.