∫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
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
For α,β,γ∈(0,1) satisfying α+β+γ=1 and
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
Notice if we substitute x by y=1−x, we have
Combine these two representations of Fαβ(μ) and let ω=(μ1−μ)γ, we obtain
Let (α,β,γ)=(14,12,14) and μ=√32, the identity we want to check becomes
Let K(m) be the complete elliptic integral of the first kind associated with modulus m. i.e.
It is known that K(12)=8π3/2Γ(−14)2. In term of K(12), it is easy to check (∗1) is equivalent to
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:
Let ψ(u) be 1√2(φ(u)+φ(u)−1). It is easy to check/verify
Compare the ODE of ψ(u) with that of a Jacobi elliptic functions with modulus m=12, we find
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.
and the condition (∗2) becomes whether following equality is true or not.
Notice 3(43K+iK)=4K+3iK is a pole of sn(u). if one repeat
apply the addition formula for sn(u+v)
One find in order for sn(3u) to blow up, sn(u) will be a root of
following polynomial equation:
Substitute m=12 and s=1√2(t+1t) into this, the equation ω need to satisfy is given by:
One can check that ω=4√√32−√3 is indeed a root of this polynomial. As a result, the original equality is valid.