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})

α=6 12+73336=(2+3)(24273)=333+2 427.
Note that α is the unique positive root of the polynomial equation
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

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

Added later: We can simplify it to

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.


Let t=1+y3, we can rewrite B(α2;12,13) as


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 ():
Let u0=i33, u1=i32 and u=i5312=12(u0+u1). Using the addition formula for function,
we have

Using the duplication formula of function, we get
Let Y=4η2(u) and A2=1Y3, above condition is equivalent to


  • 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.

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

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