# Prove ${\large\int}_0^1\frac{\ln(1+8x)}{x^{2/3}\,(1-x)^{2/3}\,(1+8x)^{1/3}}dx=\frac{\ln3}{\pi\sqrt3}\Gamma^3\!\left(\tfrac13\right)$

Here is one more numerically discovered conjecture that I was not able to prove, and asking you for help: $${\large\int}_0^1\frac{\ln(1+8x)}{x^{\small2/3}\,(1-x)^{\small2/3}\,(1+8x)^{\small1/3}}dx\stackrel{\color{#A0A0A0}{\small?}}=\frac{\ln3}{\pi\sqrt3}\Gamma^3\!\left(\tfrac13\right),\tag1$$ or, equivalently, $$\frac{d}{da}{_2F_1}\!\left(a,\frac13;\,\frac23;\,-8\right)\Bigg|_{a=\frac13}\stackrel{\color{#A0A0A0}{\small?}}=-\frac{2\ln3}3.\tag2$$

Update: I can suggest a generalization of this conjecture that might be easier to prove: $$_2F_1\!\left(-a,\frac16-\frac{a}2;\,\frac23;\,-8\right)=2\times3^{\frac{3a-1}2}\cos\left(\frac\pi6(3a+1)\right).\tag3$$ This is actually the identity 07.23.03.0658.01 from the Wolfram Functions Site generalized to non-integer values of $a$.

• This identity from the Wolfram Functions site may provide a useful starting point. (Note that the $F$ on the RHS is the so-called Kampe de Feriet function, not a generalized hypergeometric.) Jul 30, 2014 at 22:04
• Given the fact that most of your integrals are of the same form, why don't you just ask for a general approach ? Jul 30, 2014 at 22:31
• @Lucian I do not think there is a general approach. Jul 30, 2014 at 22:40
• The substitution $u=\frac{-3x\left(\sqrt[3]{1-x}\;\;_2F_1\left(\frac{1}{3},\frac{2}{3};\frac{5}{3};x\right)+2x^2-3x+1\right)}{14\sqrt[3]{-(x-1)x}}$ contains a hypergeometric function of the correct form, and has $du=(x-x^2)^{2/3}dx$, which cancels the first two terms of the denominator.
– user105475
Jul 31, 2014 at 1:33
• @user153012 Actualy, first I discovered formula $(2)$. Its right-hand side is easily recognized: isc.carma.newcastle.edu.au/… Sep 29, 2014 at 17:26

Define $\mathcal{I}$ to be the value of the definite integral,

$$\mathcal{I}:=\int_{0}^{1}\frac{\ln{\left(1+8x\right)}}{x^{2/3}(1-x)^{2/3}(1+8x)^{1/3}}\,\mathrm{d}x\approx3.8817.$$

Problem. Prove that the following conjectured value for the definite integral $\mathcal{I}$ is correct: $$\mathcal{I}=\int_{0}^{1}\frac{\ln{\left(1+8x\right)}}{x^{2/3}(1-x)^{2/3}(1+8x)^{1/3}}\,\mathrm{d}x\stackrel{?}{=}\frac{\ln{(3)}}{\sqrt{3}\,\pi}\left[\Gamma{\left(\small{\frac13}\right)}\right]^3.\tag{1}$$

Elimination of logarithmic factor from the integrand:

Suppose we have a substitution relation of the form $1+8x=\frac{k}{1+8t}$, with $k$ being some positive real constant greater than $1$. First of all, the symmetry of the relation with respect to the variables $x$ and $t$ implies that $t$ solved for as a function of $x$ will have the same functional form as $x$ solved for as a function of $t$:

$$1+8x=\frac{k}{1+8t}\implies t=\frac{k-(1+8x)}{8(1+8x)},~~x=\frac{k-(1+8t)}{8(1+8t)}.$$

Transforming the integral $\mathcal{I}$ via this substitution, we find:

\begin{align} \mathcal{I} &=\int_{0}^{1}x^{-2/3}\,(1-x)^{-2/3}\,(1+8x)^{-1/3}\,\ln{(1+8x)}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{\ln{(1+8x)}}{\sqrt[3]{x^{2}\,(1-x)^{2}\,(1+8x)}}\,\mathrm{d}x\\ &=\int_{\frac{k-1}{8}}^{\frac{k-9}{72}}\sqrt[3]{\frac{2^{12}(1+8t)^{5}}{k(9-k+72t)^{2}(k-1-8t)^{2}}}\,\ln{\left(\frac{k}{1+8t}\right)}\cdot\frac{(-k)}{(1+8t)^2}\,\mathrm{d}t\\ &=\left(\frac{k}{9}\right)^{2/3}\int_{\frac{k-9}{72}}^{\frac{k-1}{8}}\frac{\ln{\left(\frac{k}{1+8t}\right)}}{\left(\frac{9-k}{72}+t\right)^{2/3}\left(\frac{k-1}{8}-t\right)^{2/3}\left(1+8t\right)^{1/3}}\,\mathrm{d}t,\\ \end{align}

which clearly suggests the choice $k=9$ as being the simplest, in which case:

\begin{align} \mathcal{I} &=\int_{0}^{1}\frac{\ln{\left(\frac{9}{1+8t}\right)}}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\,\mathrm{d}t\\ &=\int_{0}^{1}\frac{\ln{\left(9\right)}}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\,\mathrm{d}t-\int_{0}^{1}\frac{\ln{\left(1+8t\right)}}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\,\mathrm{d}t\\ &=2\ln{(3)}\,\int_{0}^{1}\frac{\mathrm{d}t}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}-\mathcal{I}\\ \implies 2\mathcal{I}&=2\ln{(3)}\,\int_{0}^{1}\frac{\mathrm{d}t}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\\ \implies \mathcal{I}&=\ln{(3)}\,\int_{0}^{1}\frac{\mathrm{d}t}{t^{2/3}\left(1-t\right)^{2/3}\left(1+8t\right)^{1/3}}\\ &=:\ln{(3)}\,\mathcal{J}, \end{align}

where in the last line we've simply introduced the symbol $\mathcal{J}$ to denote the last integral for convenience. It's approximate value is $\mathcal{J}\approx3.53328$.

Thus, to prove that the conjectured value $(1)$ is indeed correct, it suffices to prove the following equivalent conjecture:

$$\mathcal{J}:=\int_{0}^{1}\frac{\mathrm{d}x}{x^{2/3}(1-x)^{2/3}(1+8x)^{1/3}}\stackrel{?}{=}\frac{1}{\sqrt{3}\,\pi}\left[\Gamma{\left(\small{\frac13}\right)}\right]^3.\tag{2}$$

Representation and manipulation of integral as a hypergeometric function:

Euler's integral representation for the Gauss hypergeometric function states that, for $\Re{\left(c\right)}>\Re{\left(b\right)}>0\land |\arg{\left(1-z\right)}|<\pi$, we have:

$$\int_{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\mathrm{d}x=\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)}.$$

In particular, if we choose $z=-8$, $a=\frac13$, $c=\frac23$, and $b=\frac13$, then the conditions $\Re{\left(\frac23\right)}>\Re{\left(\frac13\right)}>0\land |\arg{\left(1-(-8)\right)}|=0<\pi$ are satisfied, and the integral on the left-hand-side of Euler's representation reduces to the integral $\mathcal{J}$. That is,:

\begin{align} \mathcal{J} &=\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac23}(1+8x)^{-\frac13}\mathrm{d}x\\ &=\operatorname{B}{\left(\frac13,\frac13\right)}\,{_2F_1}{\left(\frac13,\frac13;\frac23;-8\right)}. \end{align}

$${_2F_1}{\left(a,b;2b;z\right)} = \left(\frac{1+\sqrt{1-z}}{2}\right)^{-2a} {_2F_1}{\left(a,a-b+\frac12;b+\frac12;\left(\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}}\right)^2\right)},$$

with particular values $a=b=\frac13,z=-8$, we have the hypergeometric identity,

$${_2F_1}{\left(\frac13,\frac13;\frac23;-8\right)} = 2^{-\frac23} {_2F_1}{\left(\frac13,\frac12;\frac56;\frac14\right)}.$$

Then, applying Euler's transformation,

$${_2F_1}{\left(a,b;c;z\right)}=(1-z)^{c-a-b}{_2F_1}{\left(c-a,c-b;c;z\right)},$$

we have,

$${_2F_1}{\left(\frac13,\frac12;\frac56;\frac14\right)}={_2F_1}{\left(\frac12,\frac13;\frac56;\frac14\right)}.$$

Now, Euler's integral representation for this hypergeometric function implies:

$${_2F_1}{\left(\frac12,\frac13;\frac56;\frac14\right)}=\frac{1}{\operatorname{B}{\left(\frac13,\frac12\right)}}\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac12}\left(1-\small{\frac14}x\right)^{-\frac12}\mathrm{d}x.$$

Hence,

\begin{align} \mathcal{J} &=\operatorname{B}{\left(\frac13,\frac13\right)}\,{_2F_1}{\left(\frac13,\frac13;\frac23;-8\right)}\\ &=\frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{2^{2/3}}\,{_2F_1}{\left(\frac13,\frac12;\frac56;\frac14\right)}\\ &=\frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{2^{2/3}}\,{_2F_1}{\left(\frac12,\frac13;\frac56;\frac14\right)}\\ &=\frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{2^{2/3}\,\operatorname{B}{\left(\frac13,\frac12\right)}}\,\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac12}\left(1-\small{\frac14}x\right)^{-\frac12}\mathrm{d}x.\\ \end{align}

The ratio of beta functions in the last line above simplifies considerably. The Legendre duplication formula for the gamma function states:

$$\Gamma{\left(2z\right)}=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma{\left(z\right)}\Gamma{\left(z+\frac12\right)}.$$

Letting $z=\frac13$ yields:

$$\Gamma{\left(\frac23\right)}=\frac{2^{-1/3}}{\sqrt{\pi}}\Gamma{\left(\frac13\right)}\Gamma{\left(\frac56\right)}.$$

Then, using the facts that $\operatorname{B}{\left(a,b\right)}=\frac{\Gamma{\left(a\right)}\,\Gamma{\left(b\right)}}{\Gamma{\left(a+b\right)}}$ and $\Gamma{\left(\frac12\right)}=\sqrt{\pi}$, we can simplify the ratio of beta functions above considerably:

\begin{align} \frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{\operatorname{B}{\left(\frac13,\frac12\right)}} &=\frac{\left[\Gamma{\left(\frac13\right)}\right]^2\,\Gamma{\left(\frac56\right)}}{\Gamma{\left(\frac23\right)}\,\Gamma{\left(\frac12\right)}\,\Gamma{\left(\frac13\right)}}\\ &=\frac{\Gamma{\left(\frac13\right)}\,\Gamma{\left(\frac56\right)}}{\sqrt{\pi}\,\Gamma{\left(\frac23\right)}}\\ &=\frac{\sqrt{\pi}\,\sqrt[3]{2}}{\sqrt{\pi}}\\ &=\sqrt[3]{2}. \end{align}

Thus,

\begin{align} \mathcal{J} &=\frac{\operatorname{B}{\left(\frac13,\frac13\right)}}{2^{2/3}\,\operatorname{B}{\left(\frac13,\frac12\right)}}\,\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac12}\left(1-\small{\frac14}x\right)^{-\frac12}\mathrm{d}x\\ &=\frac{\sqrt[3]{2}}{2^{2/3}}\,\int_{0}^{1}x^{-\frac23}(1-x)^{-\frac12}\left(1-\small{\frac14}x\right)^{-\frac12}\mathrm{d}x\\ &=\frac{1}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{x^{-\frac23}}{\sqrt{\left(1-x\right)\left(1-\small{\frac14}x\right)}}\,\mathrm{d}x.\\ \end{align}

Reduction of integral to pseudo-elliptic integrals:

Substituting $x=t^3$ into the integral representation for $\mathcal{J}$, we reduce the problem to solving an integral whose integrand is the reciprocal square-root of a sixth-degree polynomial:

\begin{align} \mathcal{J} &=\frac{1}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{x^{-\frac23}}{\sqrt{\left(1-x\right)\left(1-\small{\frac14}x\right)}}\,\mathrm{d}x\\ &=\frac{2}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{x^{-\frac23}}{\sqrt{\left(1-x\right)\left(4-x\right)}}\,\mathrm{d}x\\ &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{\frac13\,x^{-\frac23}\,\mathrm{d}x}{\sqrt{\left(1-x\right)\left(4-x\right)}}\\ &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{\left(1-t^3\right)\left(4-t^3\right)}}\\ &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{4-5t^3+t^6}}.\\ \end{align}

Then, substituting $t=\sqrt[3]{2}\,u$ gives us a similar integrand except with a sixth-degree polynomial with symmetric coefficients:

\begin{align} \mathcal{J} &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{4-5t^3+t^6}}\\ &=\frac{6}{\sqrt[3]{2}}\,\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{\sqrt[3]{2}\,\mathrm{d}u}{\sqrt{4-10u^3+4u^6}}\\ &=3\,\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{\mathrm{d}u}{\sqrt{1-\small{\frac52}u^3+u^6}}.\\ \end{align}

The form of the integral $\mathcal{J}$ found in the last line above is significant because it may be transformed into a pair of pseudo-elliptic integrals, which can subsequently be evaluated in terms of elementary functions and elliptic integrals. For more information on these types of integrals, see my question here.

Substituting $u=z-\sqrt{z^2-1}$ transforms the integral $\mathcal{J}$ into a sum of two pseudo-elliptic integrals:

\begin{align} \mathcal{J} &=3\,\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{\mathrm{d}u}{\sqrt{1-\small{\frac52}u^3+u^6}}\\ &=\frac{3}{\sqrt{2}}\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z+1)(8z^3-6z-\frac52)}}+\frac{3}{\sqrt{2}}\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z-1)(8z^3-6z-\frac52)}}\\ &=3\,\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z+1)(16z^3-12z-5)}}+3\,\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z-1)(16z^3-12z-5)}}\\ &=\frac34\,\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z+1)(z^3-\small{\frac34}z-\small{\frac{5}{16}})}}+\frac34\,\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z-1)(z^3-\small{\frac34}z-\small{\frac{5}{16}})}}\\ &=:\frac34\,P_{+}+\frac34\,P_{-},\\ \end{align}

where in the last line we've introduced the auxiliary notation $P_{\pm}$ to denote the pair of integrals,

$$P_{\pm}:=\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}z}{\sqrt{(z\pm1)(z^3-\small{\frac34}z-\small{\frac{5}{16}})}}.$$

Evaluation of pseudo-elliptic integrals:

Again for convenience, we shall denote the lower integration limit in the integrals $P_{\pm}$ defined above by $2^{-2/3}+2^{-4/3}=:\alpha\approx1.026811$. In particular, this eases the factorization of the cubic polynomial in the denominators above:

\begin{align} 16x^3-12x-5 &=16\left(x-\alpha\right)\left(x^2+\alpha x+\alpha^2-\small{\frac34}\right)\\ &=16\left(x-\alpha\right)\left[\left(x+\frac{\alpha}{2}\right)^2+\frac34\left(\alpha^2-1\right)\right]. \end{align}

Then,

\begin{align} P_{\pm} &=\int_{2^{-2/3}+2^{-4/3}}^{\infty}\frac{\mathrm{d}x}{\sqrt{(x\pm1)(x^3-\small{\frac34}x-\small{\frac{5}{16}})}}\\ &=\int_{\alpha}^{\infty}\frac{\mathrm{d}x}{\sqrt{(x\pm1)(x^3-\small{\frac34}x-\small{\frac{5}{16}})}}\\ &=\int_{\alpha}^{\infty}\frac{\mathrm{d}x}{\sqrt{(x\pm1)(x-\alpha)(x^2+\alpha x+\alpha^2-\small{\frac34})}}\\ &=\int_{\alpha}^{\infty}\frac{\mathrm{d}x}{\sqrt{(x\pm1)\left(x-\alpha\right)\left[\left(x+\frac{\alpha}{2}\right)^2+\small{\frac34}\left(\alpha^2-1\right)\right]}}.\\ \end{align}

EDIT: I'm beginning to fear this strategy of derivation may be ultimately fruitless. The good news is that there is a relatively compact way of expressing the two integrals above as incomplete elliptic integrals:

Assume $\alpha,\beta,u,m,n\in\mathbb{R}$ such that $\beta<\alpha<u$. Then proposition 3.145(1) of Gradshteyn's Table of Integrals, Series, and Products states: \begin{align} \small{\int_{\alpha}^{u}\frac{\mathrm{d}x}{\sqrt{\left(x-\alpha\right)\left(x-\beta\right)\left[\left(x-m\right)^2+n^2\right]}}} &=\small{\frac{1}{pq}F{\left(2\arctan{\sqrt{\frac{q\left(u-\alpha\right)}{p\left(u-\beta\right)}}},\frac12\sqrt{\frac{\left(p+q\right)^2+\left(\alpha-\beta\right)^2}{pq}}\right)},} \end{align} where $(m-\alpha)^2+n^2=p^2$, and $(m-\beta)^2+n^2=q^2$.

The bad news is after plugging in all the appropriate values, the resulting arguments of the elliptic integrals are significantly more complex than I expected.

• Thoughts on where to go from here: If we apply contour integration here we may be onto something. Perhaps a "keyhole" integral that avoids the three branch cuts at $z = re^{i\cdot \pi}$, $z = re^{i\cdot 5\pi/3}$ and $z = re^{i\cdot \pi/3}$, where $1 \leq |r| \leq 9$. The difficulty comes with the fact that the branch cut is not along the positive real axis. I suspect a transformation is in order first. Aug 2, 2014 at 1:18
• How did this answer only get 10 votes? Oct 30, 2018 at 18:46

I think I have found something for you.

See http://authors.library.caltech.edu/43489/ for the Integral Tables of the Bateman Project.

In Volume 1, p.310 (PDF p. 322) Formula (23), which I hope should be applicable for your case.

The last formula of david-h's computation is an example of a Mellin integral transformation.

$\int_0^\infty (1+x)^\nu (1+\alpha x)^\mu x^{s-1} dx = B(s,-\mu-\nu-s) \ {_2F_1}(-\mu,s;-\mu-\nu;1-\alpha)$,

with B the Beta-function, $|\text{arg}\alpha| \le \pi$ and $-\mathcal{Re}(\mu+\nu) > \mathcal{Re}(s) > 0$.