Conjecture ∫10dx3√x6√1−x√1−x(√6√12+7√3−3√3−6)2=π9(3+√24√27)\int_0^1\frac{dx}{\sqrt[3]x\,\sqrt[6]{1-x}\,\sqrt{1-x\left(\sqrt{6}\sqrt{12+7\sqrt3}-3\sqrt3-6\right)^2}}=\frac\pi9(3+\sqrt2\sqrt[4]{27})

Let
α=6 12+73336=(2+3)(24273)=333+2 427.
Note that α is the unique positive root of the polynomial equation
α4+24α3+18α227=0.
Now consider the following integral:
I=10dx3x 61x 1xα2.
I have a conjectured elementary value for it
I?=π9(3+2 427)=π3α.


Actually, the integral I can be evaluated in an exact form using Mathematica or manually, using formula DLMF 15.6.1:
I=4π3Γ(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+434427Γ(13)Γ(56)π
or
n=0Γ(n+23)(2n+1)Γ(n+1)α2n?=3+242718π3/2Γ(56).


The conjecture can also be given in terms of the incomplete beta function:
B(α2; 12,13)?=π2Γ(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+9541587503305338150665+123(165+755467503305103150665).
Added later: We can simplify it to
β=34(7+354566+305)+12495+225536(8545+38215).

I conjecture that:
10dx3x 61x 1xβ2?=2π53β.

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

α20t1/2(1t)2/3dt=31α213y2dy1+y3y2=331α21dy1+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)3g2(z)g3 where g2=0 and g3=η616
If one perform variable substitution y=4η2(izη), one has

dy1+y3=dz and {y(3η3)=4η2(i33)=0y(3η2)=4η2(i32)=1
Using this, we can express conjecture (8) in terms of y() and/or ():
B(α2;12,13)?=π2Γ(13)Γ(56)=34η3[y1(1)y1(31α2)]?=34ηy1(31α2)?=5312η4η2(i5312)?=31α2
Let u0=i33, u1=i32 and u=i5312=12(u0+u1). Using the addition formula for function,
we have

(2u)=(u0+u1)=14[(u0)(u1)(u0)(u1)]2(u0)(u1)=14[iη3400η24]20η24=η22
Using the duplication formula of function, we get
η22=(2u)=14((6(u)212g2)24(u)3g2(u)g3)2(u)=9(u)44(u)3η4162(u)
Let Y=4η2(u) and A2=1Y3, above condition is equivalent to

Y4+8Y3+8Y8=0Y(8+Y3)=8(1Y3)(9A2)3(1A2)=512A6(A424A3+18A227)(A4+24A3+18A227)=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 1α2>0, (1c)(1b) in this particular case.
    i.e. Y=31α2 satisfies (1b) and hence (1a).
  • Since u lies between u0 and u1, 4η2(u)>0. Using
    the fact (1a) has only one positive root, we find 4η2(u)=31α2. i.e. conjecture (8) is true.

Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : Community

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