∫10dx√1−x 4√x 4√2−x√3?=2√238√3π

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:

∫10dx√1−x 6√9−x 3√x?=π√3,

which is equivalent to

I(19; 16,13)?=12.

**Answer**

For α,β,γ∈(0,1) satisfying α+β+γ=1 and

μ∈C∖[1,∞), define

Fαβ(μ)=∫10dxxα(1−x)β(1−μx)γ and Δ=Γ(1−α)Γ(1−β)Γ(1+γ)

When |μ|<1, we can rewrite the integral Fαβ(μ) as

Fαβ(μ)=∫101xα(1−x)β(∞∑n=0(γ)nn!μnxn)dx=∞∑n=0(γ)nn!Γ(n+1−α)Γ(1−β)Γ(n+1+γ)μn=Δ∞∑n=0(γ)n(1−α)nn!(γ+1)nμn=Δγ∞∑n=0(1−α)nn!(γ+n)μn

This implies

μ−γ(μ∂∂μ)μγFαβ(μ)=Δγ∞∑n=0(1−α)nn!μn=Δγ1(1−μ)1−α

and hence

Fαβ(μ)=Δγμ−γ∫μ0νγ−1dν(1−ν)1−α=Δγ∫10tγ−1dt(1−μt)1−α=Δ∫10dt(1−μt1/γ)1−α

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

Fαβ(μ)=∫10dyyβ(1−y)α(1−μ−μy)γ=1(1−μ)γFβα(−μ1−μ)

Combine these two representations of Fαβ(μ) and let ω=(μ1−μ)γ, we obtain

Fαβ(μ)=Δ(1−μ)γ∫10dt(1+ω1/γt1/γ)1−β=Δμγ∫ω0dt(1+t1/γ)1−β

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

Γ(34)Γ(12)Γ(54)(√3)1/4∫ω0dt√1+t4?=2√238√3π

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

K(m)=∫10dx√(1−x2)(1−mx2)

It is known that K(12)=8π3/2Γ(−14)2. In term of K(12), it is easy to check (∗1) is equivalent to

∫ω0dt√1+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)0dt√1+t4

Let ψ(u) be 1√2(φ(u)+φ(u)−1). It is easy to check/verify

φ′(u)2=1+φ(u)4⟹ψ′(u)2=4(1−ψ(u)2)(1−12ψ(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.

1√2(ω+ω−1)?=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)1−msn(u)2sn(v)2

One find in order for sn(3u) to blow up, sn(u) will be a root of

following polynomial equation:

3m2s8−4m2s6−4ms6+6ms4−1=0

Substitute m=12 and s=1√2(t+1t) into this, the equation ω need to satisfy is given by:

(t8−6t4−3)(3t8+6t4−1)=0

One can check that ω=4√√32−√3 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*