Conjecture ∫10dx√1−x 4√x 4√2−x√3?=2√238√3π\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi

10dx1x 4x 42x3?=22383π
The equality numerically holds up to at least 104 decimal digits.

Can we prove that the equality is exact?

An equivalent form of this conjecture is
I(32; 14,14)?=23,
where I(z; a,b) is the regularized beta function.


Even simpler case:
10dx1x 69x 3x?=π3,
which is equivalent to
I(19; 16,13)?=12.


A related question.

Answer

For α,β,γ(0,1) satisfying α+β+γ=1 and
μC[1,), define

Fαβ(μ)=10dxxα(1x)β(1μx)γ and Δ=Γ(1α)Γ(1β)Γ(1+γ)
When |μ|<1, we can rewrite the integral Fαβ(μ) as

Fαβ(μ)=101xα(1x)β(n=0(γ)nn!μnxn)dx=n=0(γ)nn!Γ(n+1α)Γ(1β)Γ(n+1+γ)μn=Δn=0(γ)n(1α)nn!(γ+1)nμn=Δγn=0(1α)nn!(γ+n)μn
This implies
μγ(μμ)μγFαβ(μ)=Δγn=0(1α)nn!μn=Δγ1(1μ)1α
and hence
Fαβ(μ)=Δγμγμ0νγ1dν(1ν)1α=Δγ10tγ1dt(1μt)1α=Δ10dt(1μt1/γ)1α

Notice if we substitute x by y=1x, we have

Fαβ(μ)=10dyyβ(1y)α(1μμy)γ=1(1μ)γFβα(μ1μ)

Combine these two representations of Fαβ(μ) and let ω=(μ1μ)γ, we obtain

Fαβ(μ)=Δ(1μ)γ10dt(1+ω1/γt1/γ)1β=Δμγω0dt(1+t1/γ)1β

Let (α,β,γ)=(14,12,14) and μ=32, the identity we want to check becomes

Γ(34)Γ(12)Γ(54)(3)1/4ω0dt1+t4?=22383π

Let K(m) be the complete elliptic integral of the first kind associated with modulus m. i.e.

K(m)=10dx(1x2)(1mx2)
It is known that K(12)=8π3/2Γ(14)2. In term of K(12), it is easy to check (1) is equivalent to

ω0dt1+t4?=23K(12)

To see whether this is the case, let φ(u) be the inverse function of above integral.
More precisely, define φ(u) by following relation:

u=φ(u)0dt1+t4

Let ψ(u) be 12(φ(u)+φ(u)1). It is easy to check/verify
φ(u)2=1+φ(u)4ψ(u)2=4(1ψ(u)2)(112ψ(u)2)

Compare the ODE of ψ(u) with that of a Jacobi elliptic functions with modulus m=12, we find

ψ(u)=sn(2u+constant|12)

Since we are going to deal with elliptic functions/integrals with m=12 only,
we will simplify our notations and drop all reference to modulus, i.e
sn(u) now means sn(u|m=12) and K means K(m=12).

Over the complex plane, it is known that sn(u) is doubly periodic with
fundamental period 4K and 2iK. It has two poles at iK and (2+i)K in the fundamental domain.
When u=0, we want φ(u)=0 and hence ψ(u)=. So the constant
in (3) has to be one of the pole. For small and positive u, we want φ(u) and hence ψ(u) to be positive. This fixes the constant to iK. i.e.

ψ(u)=sn(2u+iK)

and the condition (2) becomes whether following equality is true or not.

12(ω+ω1)?=sn(43K+iK)

Notice 3(43K+iK)=4K+3iK is a pole of sn(u). if one repeat
apply the addition formula for sn(u+v)

sn(u+v)=sn(u)cn(v)dn(v)+sn(v)cn(u)dn(u)1msn(u)2sn(v)2

One find in order for sn(3u) to blow up, sn(u) will be a root of
following polynomial equation:
3m2s84m2s64ms6+6ms41=0
Substitute m=12 and s=12(t+1t) into this, the equation ω need to satisfy is given by:

(t86t43)(3t8+6t41)=0

One can check that ω=4323 is indeed a root of this polynomial. As a result, the original equality is valid.

Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : achille hui

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