18
$\begingroup$

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 $a$.

$\endgroup$
  • $\begingroup$ 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.) $\endgroup$ – Semiclassical Jul 30 '14 at 22:04
  • 1
    $\begingroup$ Given the fact that most of your integrals are of the same form, why don't you just ask for a general approach ? $\endgroup$ – Lucian Jul 30 '14 at 22:31
  • 3
    $\begingroup$ @Lucian I do not think there is a general approach. $\endgroup$ – Vladimir Reshetnikov Jul 30 '14 at 22:40
  • $\begingroup$ 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. $\endgroup$ – user105475 Jul 31 '14 at 1:33
  • 1
    $\begingroup$ @user153012 Actualy, first I discovered formula $(2)$. Its right-hand side is easily recognized: isc.carma.newcastle.edu.au/… $\endgroup$ – Vladimir Reshetnikov Sep 29 '14 at 17:26
11
$\begingroup$

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,:

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

Using the quadratic transformation,

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

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Chris K Aug 2 '14 at 1:18
  • $\begingroup$ How did this answer only get 10 votes? $\endgroup$ – clathratus Oct 30 '18 at 18:46
6
$\begingroup$

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 $-\mathcal{Re}(\mu+\nu) > \mathcal{Re}(s) > 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.