Let
α=√6 √12+7√3−3√3−6=(2+√3)(√24√27−3)=3√33+√2 4√27.
Note that α is the unique positive root of the polynomial equation
α4+24α3+18α2−27=0.
Now consider the following integral:
I=∫10dx3√x 6√1−x √1−xα2.
I have a conjectured elementary value for it
I?=π9(3+√2 4√27)=π√3α.
Actually, the integral I can be evaluated in an exact form using Mathematica or manually, using formula DLMF 15.6.1:
I=4√π√3⋅Γ(56)Γ(13)⋅2F1(12,23; 32; α2),
but I could not find a way to simplify this result to (4).So, my conjecture can be restated in a different form:
2F1(12,23; 32; α2)?=√2+4√344√27⋅Γ(13)Γ(56)⋅√π
or
∞∑n=0Γ(n+23)(2n+1)Γ(n+1)α2n?=3+√24√2718⋅π3/2Γ(56).
The conjecture can also be given in terms of the incomplete beta function:
B(α2; 12,13)?=√π2⋅Γ(13)Γ(56).
Question: Is the conjecture indeed true?
Note: It holds numerically up to at least 104 decimal digits.
Conjecture 2
Let
β=214+9√54−158√750−330√5−338√150−66√5+12√3(165+75√5−46√750−330√5−103√150−66√5).
Added later: We can simplify it to
β=34(7+3√5−4√5√66+30√5)+12√495+225√5−3√6(8545+3821√5).I conjecture that:
∫10dx3√x 6√1−x √1−xβ2?=2π5√3β.I can imagine that if these conjectures are true, then there are generalizations for some other algebraic numbers.
Answer
Let t=1+y3, we can rewrite B(α2;12,13) as
∫α20t−1/2(1−t)−2/3dt=∫−3√1−α2−13y2dy√1+y3y2=3∫−3√1−α2−1dy√1+y3
Following the setup in my answer to a related question. Let
η=Γ(13)Γ(16)√3π and ℘(z) be the Weierstrass elliptic ℘ function with fundamental periods 1 and eiπ/3. ℘(z) is known to satisfy an ODE of the form
℘′(z)2=4℘(z)3−g2℘(z)−g3 where g2=0 and g3=η616
If one perform variable substitution y=−4η2℘(izη), one has
dy√1+y3=−dz and {y(√3η3)=−4η2℘(i√33)=0y(√3η2)=−4η2℘(i√32)=−1
Using this, we can express conjecture (8) in terms of y(⋅) and/or ℘(⋅):
B(α2;12,13)?=√π2Γ(13)Γ(56)=√34η⟺3[y−1(−1)−y−1(−3√1−α2)]?=√34η⟺y−1(−3√1−α2)?=5√312η⟺4η2℘(i5√312)?=3√1−α2
Let u0=i√33, u−1=i√32 and u=i5√312=12(u0+u−1). Using the addition formula for ℘ function,
we have
℘(2u)=℘(u0+u−1)=14[℘′(u0)−℘′(u−1)℘(u0)−℘(u−1)]2−℘(u0)−℘(u−1)=14[−iη34−00−η24]2−0−η24=−η22
Using the duplication formula of ℘ function, we get
−η22=℘(2u)=14((6℘(u)2−12g2)24℘(u)3−g2℘(u)−g3)−2℘(u)=9℘(u)44℘(u)3−η416−2℘(u)
Let Y=4η2℘(u) and A2=1−Y3, above condition is equivalent to
Y4+8Y3+8Y−8=0⟺Y(8+Y3)=8(1−Y3)⟹(9−A2)3(1−A2)=512A6⟺(A4−24A3+18A2−27)(A4+24A3+18A2−27)=0
- Since α is a root for one of the factors in (∗1d), A=α satisfies (∗1d) and hence (∗1c).
- Since 0<α<1 implies 1−α2>0, (∗1c)⟹(∗1b) in this particular case.
i.e. Y=3√1−α2 satisfies (∗1b) and hence (∗1a). - Since u lies between u0 and u−1, 4η2℘(u)>0. Using
the fact (∗1a) has only one positive root, we find 4η2℘(u)=3√1−α2. i.e. conjecture (8) is true.
Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : Community