∫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