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My brother asked me to calculate the following integral before we had dinner and I have been working to calculate it since then ($\pm\, 4$ hours). He said, it has a beautiful closed form but I doubt it and I guess he has tried to trick me again (as usual). So I am curious, what is the closed form (if any) of the following integral:

\begin{equation} \int_{-1}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\,dx \end{equation}

I have tried by parts method, partial fractions (stupid idea), converting into series (nothing familiar), many substitutions such as: $u=2x-1$, $u=1-x$, $x=\cos^2\theta$, etc, but I failed and got nothing. Wolfram Alpha also doesn't give an answer. Either he is lying to me or telling the truth, I don't know. Could anyone here please help me to obtain the closed form of the integral with any methods (whatever it takes)? Any help would be greatly appreciated. Thank you.

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    $\begingroup$ What is the ln of a negative number? $\endgroup$ – André Nicolas Jul 26 '14 at 15:46
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    $\begingroup$ Didn't your mother ever warn you not to do integrals before dinner? $\endgroup$ – copper.hat Jul 26 '14 at 16:07
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    $\begingroup$ If you restrict the interval of integral to $[\frac12,1]$ where the integrand is real valued, the substitution $x(1-x)=u$ transforms the integral into $\int_{0}^{\frac14}\frac{\ln{\sqrt{1-4u}}}{u^{1/6}(1-4u)^{5/6}}\,du$, which is just a hop, skip and u-substitution away from a derivative of a beta function. Perhaps this is what your brother noticed??? $\endgroup$ – David H Jul 26 '14 at 16:15
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    $\begingroup$ The integral over $[\frac12,1]$ I mentioned has a nice closed form: $I=-\frac{\pi}{2^{8/3}}\left(\sqrt{3}\,\pi+\ln{432}\right)$. $\endgroup$ – David H Jul 26 '14 at 16:37
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Too long for a comment and this is not the answer too, but following @David H's comment. Perhaps your brother means

$$\int_{\color{red}{\large\frac12}}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\ dx.\tag1$$

If so, then setting $u=2x-1$ to $(1)$ yields $$ \frac1{\sqrt[\large 3]{4}}\int_{0}^1\frac{\ln u}{\sqrt[\large 6]{(1+u)(1-u)u^4}}\ du=\frac1{\sqrt[\large 3]{4}}\int_{0}^1\frac{\ln u}{\sqrt[\large 6]{(1-u^2)u^4}}\ du.\tag2 $$ Setting $t=u^2$ yields $$ \frac1{\sqrt[\large 3]{128}}\int_{0}^1\frac{\ln t}{(1-t)^{\large\frac16}t^{\large\frac56}}\ dt=\frac1{\sqrt[\large 3]{128}}\int_{0}^1(1-t)^{-\large\frac16}t^{\large-\frac56}\ln t\ dt.\tag3 $$

Now, consider beta function $$ \text{Beta}(x,y)=\int_{0}^1t^{\large x-1}(1-t)^{\large y-1}\ dt.\tag4 $$ Differentiating $(4)$ with respect to $x$ yields \begin{align} \frac{\partial}{\partial x}\text{B}(x,y)&=\int_0^1\frac{\partial}{\partial x}\left(t^{\large x-1}(1-t)^{\large y-1}\right)\ dt\\ (\psi(x)-\psi(x+y))\text{B}(x,y)&=\int_0^1 t^{\large x-1}(1-t)^{\large y-1}\ln t\ dt,\tag5 \end{align} where $\psi(\cdot)$ is the digamma function.

Using $(5)$ then $(3)$ turns out to be $$ \frac1{\sqrt[\large 3]{128}}\left(\psi\left(\frac16\right)-\psi\left(1\right)\right)\text{B}\left(\frac16,\frac56\right)=\large\color{blue}{-\frac\pi{\sqrt[\large 3]{256}}\left(\pi\sqrt{3}+4\ln2+3\ln3\right)}. $$ Either your brother is teasing you or he just misplaces the interval of integral.

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    $\begingroup$ I like how you substituted $u=1-2*x$ instead of $x(1-x)$ like I chose to, because it demonstrates the underlying symmetry of the integral in these factors. This result is pleasing enough to me to give the brother the benefit of the doubt. +1 $\endgroup$ – David H Jul 26 '14 at 17:26
  • $\begingroup$ @DavidH Thanks. Anyway, I like your answer here. I tried using derivative of beta function to obtain the integral but I couldn't. +1 $\endgroup$ – Tunk-Fey Jul 26 '14 at 17:36
  • $\begingroup$ Now that you mention it, you do appear to have an incorrect final result. You can verify numerically the integral is approximately $\approx -5.6947...$, but your value comes to $-28.69956...$. Somewhere you lost a factor of $2^{-7/3}$. $\endgroup$ – David H Jul 26 '14 at 17:45
  • $\begingroup$ And thank you for the compliment on that answer other you mentioned. I tried and failed to create an answer to this question modeled off of that one, and had to resort to more involved methods. $\endgroup$ – David H Jul 26 '14 at 17:49
  • $\begingroup$ @DavidH Thanks for pointing that out. I've edited my answer. Now it numerically equals. $\endgroup$ – Tunk-Fey Jul 26 '14 at 18:06
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I highly doubt that one can evaluate that integral by hand, however the integral does have a closed form found by Mathematica to be: $$\frac{2 \left(4 (-1)^{5/6} \, _4F_3\left(1,1,1,\frac{7}{6};2,2,2;\frac{1}{9}\right)+8 (-1)^{5/6} \log (3) \, _3F_2\left(1,1,\frac{7}{6};2,2;\frac{1}{9}\right)+432 (-1)^{5/6} \log ^2(2)-189 (-1)^{5/6} \log ^2(3)+648 (-1)^{5/6} \log (3) \log (2)-54 i \psi ^{(1)}\left(\frac{5}{6}\right)+54 \sqrt{3} \psi ^{(1)}\left(\frac{5}{6}\right)\right)+\pi ^2 \left(-963 \sqrt{3}+1827 i\right)-54 i \left(\sqrt{3}-13 i\right) \pi \log (432)}{1728\ 2^{2/3}},$$ where $pF_q(a;b;z)$ is the Generalized Hypergeometric function, and $\psi^{(n)}$ is the Polygamma function of order $n$.

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