Let K(x) and E(x) denote the complete elliptic integrals of the first and second kinds, respectively, and let K'(x) and E'(x) denote their complementary counterparts. The integrals are taken over 0≤ x≤ 1. Elliptic-integral moment conjecture. int_0^1 (2K(x)^2-4E(x)K(x)+3E(x)^2-K'(x)E'(x)),dx=0. The paper states that this is the only entry in its tables for which the authors do not possess a proof, so its status remains open.

Unproved elliptic-integral moment identity involving derivatives

From papers

Let $K(x)$ and $E(x)$ denote the complete elliptic integrals of the first and second kinds, respectively, and let $K'(x)$ and $E'(x)$ denote their complementary counterparts. The integrals are taken over $0\leq x\leq 1$ . Elliptic-integral moment conjecture.

\int_0^1 \left(2K(x)^2-4E(x)K(x)+3E(x)^2-K'(x)E'(x)\right)\,\mathrm{d}x=0.

The paper states that this is the only entry in its tables for which the authors do not possess a proof, so its status remains open.

Progress summary

Open

The identity remains an open conjecture, with no publicly verified proof or disproof found.

The identity was posed in the 2011 paper Moments of Products of Elliptic Integrals. The authors stated that it was the sole unproved entry in their tables and that proving it would determine five related moments, while they had derived only four relations.

Known results

The 2011 paper derives four relations among the five relevant moments, leaving this identity as the missing relation.
A 2013 follow-up settles several conjectures from the earlier paper and gives additional elliptic-integral moment evaluations, but does not identify this specific identity as settled.

Current status (as of August 2026): The conjectured identity remains open; no verified proof, disproof, or claim specifically resolving it was found.

Sources

ar5iv.labs.arxiv.org

Sources & referencesView supporting material

Primary source

James Wan, “Moments of Products of Elliptic Integrals”, arXiv:1101.1132 (2011).

elliptic-integrals special-functions

1 week ago

Proof

Here the primes denote complementary modulus, not differentiation:

$K'(x)=K(\sqrt{1-x^2})$ and $E'(x)=E(\sqrt{1-x^2})$ .

First consider the complementary moment. Make the substitution

$t=(1-x)/(1+x)$ ,

so that

$x=(1-t)/(1+t)$ , $\quad dx=-2\,dt/(1+t)^2$ ,

and

$\sqrt{1-x^2}=2\sqrt{t}/(1+t)$ .

The Landen formulas give

$K(2\sqrt{t}/(1+t))=(1+t)K(t)$

and

$E(2\sqrt{t}/(1+t)) =\bigl(2E(t)-(1-t^2)K(t)\bigr)/(1+t)$ .

Consequently,

\int_0^1 K'(x)E'(x)\,dx = \int_0^1 \left( \frac{4K(t)E(t)}{(1+t)^2} -\frac{2(1-t)}{1+t}K(t)^2 \right)dt.

Subtracting this from the other three terms in the conjectured integral and renaming $t$ as $x$ , it remains to prove

\int_0^1 F(x)\,dx=0,

where

F(x)= \frac{4}{1+x}K(x)^2 -\left(4+\frac{4}{(1+x)^2}\right)K(x)E(x) +3E(x)^2.

Use the standard derivative identities

\frac{dK}{dx} = \frac{E-(1-x^2)K}{x(1-x^2)}, \qquad \frac{dE}{dx} = \frac{E-K}{x}.

Define

G(x)=(x-1)\left( 2K(x)^2 -2\frac{x+2}{x+1}K(x)E(x) +\frac{x+2}{x+1}E(x)^2 \right).

Direct differentiation using the two identities above, followed by collecting the coefficients of $K^2$ , $KE$ , and $E^2$ , gives

\frac{dG}{dx}=F(x).

At $x=0$ , since $K(0)=E(0)=\pi/2$ ,

G(0)=-2\bigl(K(0)-E(0)\bigr)^2=0.

As $x\to1^-$ , we have $E(x)\to1$ and

K(x)=O\left(\log\frac1{1-x}\right).

The factor $x-1$ in $G$ therefore implies $G(x)\to0$ , because

(1-x)\log^2\frac1{1-x}\longrightarrow0.

The same estimates show that the improper integrals converge. Applying the fundamental theorem of calculus first on compact subintervals and then passing to the endpoints gives

\int_0^1F(x)\,dx=G(1^-)-G(0)=0.

Together with the Landen substitution, this proves

\int_0^1 \left( 2K(x)^2-4E(x)K(x)+3E(x)^2-K'(x)E'(x) \right)dx=0.

Sources:

Original conjecture: https://arxiv.org/abs/1101.1132

Landen transformation: https://dlmf.nist.gov/19.8.E12

Derivative identities: https://dlmf.nist.gov/19.4.E1 https://dlmf.nist.gov/19.4.E2

Endpoint asymptotics: https://dlmf.nist.gov/19.12

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Samuel Schlesinger · 4 days ago