8

Background

let $\displaystyle I_k=\int_{0}^{1} x^k K(x) E(x) \,\mathrm{d} x$.

where $K(z)$ is the complete elliptic integral of the 1st kind, $E(z)$ is the complete elliptic integral of the 2nd kind.

By numerical calculation I found:

$$I_0=\int_{0}^{1} K(x) E(x) \,\mathrm{d} x\overset{?}{=}\frac{1}{2}+\frac{7}{4} \zeta(3)$$

$$I_1=\int_{0}^{1} x K(x) E(x) \,\mathrm{d} x\overset{?}{=}\frac{9}{16}+\frac{21}{32} \zeta(3)$$

Numerical verification:

NIntegrate[EllipticK[x] EllipticE[x], {x, 0, 1}, WorkingPrecision -> 100]
N[(7 Zeta[3] + 2)/4, 100]
NIntegrate[x EllipticK[x] EllipticE[x], {x, 0, 1}, WorkingPrecision -> 100]
N[9 / 16 + 21 Zeta[3] / 32, 100]

Is it possible to prove this conclusion?

Update 20220521

This article gives really helpful results: https://arxiv.org/pdf/1101.1132.pdf

On page7, the authors give formula of K'E', KE, K'E, KE', but without KE.

The most closed formula is K'E' in page7

$$ \int_0^1 x^n K\left(1-x^2\right) E\left(1-x^2\right) \, dx=\frac{2^{4 n}(n+1)^{3}(n+3)}{16(n+2)^{3}} \frac{\Gamma\left(\frac{1}{2}(n+1)\right)^{8}}{\Gamma(n+1)^{4}}{ }_{7} F_{6}\left(\begin{array}{c} \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{n+3}{2}, \frac{n+3}{2}, \frac{n+7}{4} \\ 1, \frac{n+3}{4}, \frac{n+2}{2}, \frac{n+4}{2}, \frac{n+4}{2}, \frac{n+4}{2} \end{array} \mid 1\right)$$

let $n = 1$, then:

$$S = \int_0^1 x K\left(1-x^2\right) E\left(1-x^2\right) \, dx$$

let $t = 1-x^2$, then:

$$S = \frac{1}{2} \int_0^1 K(x) E(x) \, dx = \frac{1}{8} (7 \zeta (3)+2)$$

which gives $I_0$, but I can't figure out how to get $I_1$.

Numerical verification of hypergeometric functions given in the paper:

NIntegrate[x^2 EllipticK[1-x^2]EllipticE[1-x^2] ,{x,0,1}]
NIntegrate[Sqrt[1-x] EllipticK[x]EllipticE[x] ,{x,0,1}]/2

Clear[n, p, q]
n = 2.;
NIntegrate[x^n EllipticK[1 - x^2]EllipticE[1 - x^2] , {x, 0, 1}]
p = 2^(4n) (n + 1)^2Gamma[(n + 1) / 2]^8 / (16(n + 2)Gamma[n + 1]^4);
q = HypergeometricPFQ[
    {-1 / 2, 1 / 2, 1 / 2, 1 / 2, (n + 1) / 2, (n + 1) / 2, (n + 5) / 4},
    {1, (n + 1) / 4, n / 2 + 1, n / 2 + 1, n / 2 + 1, n / 2 + 2},
    1
];
p q
p = 2^(4n) (n + 1)^3(n + 3)Gamma[(n + 1) / 2]^8 / (16(n + 2)^3Gamma[n + 1]^4);
q = HypergeometricPFQ[
    {1 / 2, 1 / 2, 1 / 2, 3 / 2, (n + 3) / 2, (n + 3) / 2, (n + 7) / 4},
    {1, (n + 3) / 4, n / 2 + 1, n / 2 + 2, n / 2 + 2, n / 2 + 2},
    1
];
p q
user170231
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Aster
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    Why should this question be closed ? One can, in a first step, encourage the asker to explain a little more what he/she has done, where this issue is coming from, etc... – Jean Marie May 21 '22 at 05:55
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    Look at this very close question – Jean Marie May 21 '22 at 06:13
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    Agreed with you @JeanMarie, questions like this don't need to be closed right away. (1) It is interesting and (2) the author hasn't been given a chance to clarify the context. – Stefan Lafon May 21 '22 at 06:24
  • Which argument convention are you using for the elliptic integrals? Elliptic modulus or parameter? – David H May 24 '22 at 02:34
  • For generalization: $$\int_0^1 x^r K(b x) E(a x) , dx=\sum {n=0}^{\infty } \sum _{m=0}^{\infty } \frac{\pi ^2 \Gamma (1+r) \left(\left(-\frac{1}{2}\right)_n \left(\left(\frac{1}{2}\right)_m\right){}^2 \left(\frac{1}{2}\right)_n (1+r){m+n}\right) \left(a^n b^m\right)}{4 \Gamma (2+r) \left((2+r)_{m+n} (1)_n (1)_m\right) (n! m!)}=\text{KdF}\left({1,0,2};{1,1,1},\left( \begin{array}{ccc} {1+r} & {} & \left{-\frac{1}{2},\frac{1}{2}\right} \ {2+r} & {1} & {1} \ \end{array} \right),{b,a}\right)$$ where $KdF$ is Kampé de Fériet function. – Mariusz Iwaniuk Dec 03 '24 at 09:43

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