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I was trying do variations of an integral representation for $\log\frac{\pi}{2}$ due to Jonathan Sondow, when I am wondering about if it is possible to evaluate

$$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx,\tag{1}$$ Wolfram Alpha online calculator provide me a closed-form with code

int -(1-x)/((1-x+x^2)log(x)) dx, from x=0 to x=1

Question. Please provide me hints to know how evaluate previous this definite integral as $$\int_0^1-\frac{1-x}{(1-x+x^2)\log x}\,dx=\log \left(\frac{\Gamma(1/6)}{\Gamma(2/3)}\right)-\frac{\log \pi}{2}$$ as said Wolfram Alpha. Many thanks.

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    $\begingroup$ Maybe this might be helpful: $x^2-x+1=\frac{x^3+1}{x+1}$ $\endgroup$ – Simply Beautiful Art Jul 16 '17 at 12:29
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    $\begingroup$ i think this would make the question much better... $\endgroup$ – tired Jul 16 '17 at 12:34
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    $\begingroup$ $$2I'_{a/3}=\sum_{n\geq0}\frac{(-1)^n}{n+a/3}=\psi(1-\frac a6)-\psi(\frac12-\frac a6)$$ and likewise $$ I'_{a/3+2/3}=\sum_{n\geq0}\frac{(-1)^n}{n+a/3}=\psi(\frac 13-\frac a6)-\psi(\frac16-\frac a6) $$ where $\psi(z) =\log(\Gamma(z))'$ is the digamma function [the digamma function][1]. Integrating with respect to $a$ therefore yields $$6J_a=\log(\Gamma(1-\frac a6))-\log(\Gamma(\frac12-\frac a6))-\log(\Gamma(\frac13-\frac a6))+\log(\Gamma(\frac16-\frac a6))+C$$ Now your integral in question is just $J_1$... [1]: en.wikipedia.org/wiki/Digamma_function $\endgroup$ – tired Jul 16 '17 at 13:11
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    $\begingroup$ @tired Or you just let $t=e^x$, and apply some complex analysis. $\endgroup$ – Simply Beautiful Art Jul 16 '17 at 13:11
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    $\begingroup$ Please don't use int -(1-x)/((1-x+x^2)log(x)) dx, from x =0 to x=1 to replace $$\int_0^1 -\frac{(1-x)}{(1-x + x^2)\log(x)} = \int_0^1 \frac{x-1}{(x^2 -x+1)\log(x)}$$ $\endgroup$ – Namaste Jul 16 '17 at 16:13
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Thanks to the power series representation of $1/(1+x^3)$ and the Dominated Convergence Theorem, the given integral is \begin{align*} -\int_0^1\frac{1-x^2}{(1+x^3)\log x}\,dx &=\int_0^1\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(1-x^2)x^{3k}}{\log x}\,dx\\ &=\int_0^1\sum_{k=0}^{\infty}\frac{(1-x^2)(x^{6k+3}-x^{6k})}{\log x}\,dx\\ &=\sum_{k=0}^{\infty}\int_0^1\frac{(1-x^2)(x^{6k+3}-x^{6k})}{\log x}\,dx\\&= \sum_{k=0}^{\infty}(\ln(6k+3)-\ln(6k+1)+\ln(6k+4)-\ln(6k+6))\\&= \ln\left(\prod_{k=0}^{\infty}\frac{(6k+3)(6k+4)}{(6k+1)(6k+6)}\right) =\ln\left(\frac{\Gamma(1/6)}{\sqrt{\pi}\Gamma(2/3)}\right) \end{align*} where $\Gamma(x)$ is the Gamma function and we used the fact that $$\int_{0}^{1}\frac{x^n-1}{\log x}\,dx = \int_{0}^{+\infty}\frac{1-e^{-nt}}{t}\,e^{-t}dt = \log(n+1)$$ (apply Frullani's integral in the last step).

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    $\begingroup$ @Zophikel Yes to interchange the sum and the integral I used Dominated Convergence $\endgroup$ – Robert Z Jul 16 '17 at 12:59
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    $\begingroup$ Although I am certain that your computation can be justified, I would like to mention that the Fubini's theorem does not apply directly since $$ \int_{0}^{1} \sum_{k=0}^{\infty} \left| (-1)^{k+1} \frac{(1-x^2)x^{3k}}{\log x} \right| \, dx = \infty. $$ You may need more elaborated technique to justify that step, for instance regularizing the denominator and using a combination of both Fubini's theorem and the dominated convergence theorem. $\endgroup$ – Sangchul Lee Jul 16 '17 at 13:00
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    $\begingroup$ Hmmmm,,, now looking back at the question maybe there's a simpler way via Asymptotics $\endgroup$ – Zophikel Jul 16 '17 at 13:03
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    $\begingroup$ I also tried a similar approach as @tired. I computed the integral $$ I(s, a) := \int_{0}^{\infty} \frac{t^{s-1}e^{-at}}{1+e^{-t}} \, dt = \frac{\Gamma(s)}{2^s} \left[ \zeta\left(s, \frac{a}{2}\right) - \zeta\left(s, \frac{a+1}{2}\right) \right] $$ and then tried to use a functional equation for Hurwitz zeta function to compute the limit $$ \int_{0}^{1} \frac{1-x}{(1-x+x^2)\log(1/x)} \, dx = \lim_{s\to 0} [ I(s, \tfrac{1}{3}) - I(s, 1) ]. $$ But your approach is much neater and simpler. $\endgroup$ – Sangchul Lee Jul 16 '17 at 13:28
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    $\begingroup$ @SangchulLee yeah the limiting procedure becomes quite daunting :( $\endgroup$ – tired Jul 16 '17 at 13:29
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I was confused by the implicit use of the Frullani integral, so I think it bears mention: $$ \begin{align} \int_0^1\frac{x^n-x^m}{\log(x)}\,\mathrm{d}x &=\int_0^\infty\frac{e^{-mu}-e^{-nu}}{u}e^{-u}\,\mathrm{d}u\\ &=\lim_{\epsilon\to0^+}\int_\epsilon^\infty\frac{e^{-(m+1)u}-e^{-(n+1)u}}{u}\,\mathrm{d}u\\ &=\lim_{\epsilon\to0^+}\int_{(m+1)\epsilon}^{(n+1)\epsilon}\frac{e^{-u}}{u}\,\mathrm{d}u\\[3pt] &=\log\left(\frac{n+1}{m+1}\right) \end{align} $$ $$ \begin{align} -\int_0^1\frac{1-x}{\left(1-x+x^2\right)\log(x)}\,\mathrm{d}x &=-\int_0^1\frac{1-x^2}{\left(1+x^3\right)\log(x)}\,\mathrm{d}x\\ &=-\sum_{k=0}^\infty(-1)^k\frac{x^{3k}-x^{3k+2}}{\log(x)}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\log\left(\frac{3k+3}{3k+1}\right)\\ &=\sum_{k=0}^\infty\log\left(\frac{(6k+3)(6k+4)}{(6k+1)(6k+6)}\right)\\ &=\log\left(\prod_{k=0}^\infty\frac{\left(k+\frac12\right)\left(k+\frac23\right)}{\left(k+\frac16\right)(k+1)}\right)\\ &=\lim_{n\to\infty}\log\left(\prod_{k=0}^{n-1}\frac{\color{#C00}{\left(k+\frac12\right)}\color{#090}{\left(k+\frac23\right)}}{\color{#00F}{\left(k+\frac16\right)}(k+1)}\right)\\ &=\log\left(\lim_{n\to\infty}\color{#C00}{\frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)}}\color{#090}{\frac{\Gamma\left(n+\frac23\right)}{\Gamma\left(\frac23\right)}}\color{#00F}{\frac{\Gamma\left(\frac16\right)}{\Gamma\left(n+\frac16\right)}}\frac{\Gamma(1)}{\Gamma(n+1)}\right)\\ &=\log\left(\frac1{\sqrt\pi}\frac{\Gamma\left(\frac16\right)}{\Gamma\left(\frac23\right)}\right)+\log\left(\lim_{n\to\infty}\frac{\Gamma\left(n+\frac12\right)\Gamma\left(n+\frac23\right)}{\Gamma\left(n+\frac16\right)\Gamma(n+1)}\right)\\ &=\log\left(\frac1{\sqrt\pi}\frac{\Gamma\left(\frac16\right)}{\Gamma\left(\frac23\right)}\right) \end{align} $$ The last step is by Gautschi's Inequality.

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  • $\begingroup$ Many thanks for your details, and approach using the inequality. I am going to study it. Many thanks all users, those that were interested in this question, especially who provided an early answer. $\endgroup$ – user243301 Jul 16 '17 at 19:58
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &-\int_{0}^{1}{1 - x \over \pars{1 - x + x^{2}}\ln\pars{x}}\,\dd x = \int_{0}^{1}\ \overbrace{{1 + x \over 1 + x^{3}}} ^{\ds{1 \over 1 - x + x^{2}}}\,\,\, \overbrace{{x - 1 \over \ln\pars{x}}}^{\ds{\int_{0}^{1}x^{t}\,\dd t}}\ \,\dd x = \int_{0}^{1}\int_{0}^{1}{x^{t} + x^{t + 1} \over 1 + x^{3}}\,\dd x\,\dd t \\[5mm] & = \int_{0}^{1}\int_{0}^{1}{x^{t} + x^{t + 1} - x^{t + 3} - x^{t + 4} \over 1 - x^{6}}\,\dd x\,\dd t \\[5mm] & \stackrel{x^{6}\ \mapsto\ x}{=}\,\,\, {1 \over 6}\int_{0}^{1}\int_{0}^{1} {x^{t/6 - 5/6} + x^{t/6 - 2/3} - x^{t/6 - 1/3} - x^{t/6 - 1/6} \over 1 - x}\,\dd x\,\dd t \\[5mm] & = {1 \over 6}\int_{0}^{1}\!\!\pars{% -\int_{0}^{1}\!\!{1 - x^{t/6 - 5/6} \over 1 - x}\,\dd x - \int_{0}^{1}\!\!{1 - x^{t/6 - 2/3} \over 1 - x}\,\dd x + \int_{0}^{1}\!\!{1 - x^{t/6 - 1/3} \over 1 - x}\,\dd x + \int_{0}^{1}\!\!{1 - x^{t/6 - 1/6} \over 1 - x}\,\dd x} \\[5mm] & = {1 \over 6}\ \overbrace{\int_{0}^{1}\bracks{% -\Psi\pars{{t \over 6} + {1 \over 6}} - \Psi\pars{{t \over 6} + {1 \over 3}} + \Psi\pars{{t \over 6} + {2 \over 3}} + \Psi\pars{{t \over 6} + {5 \over 6}}}\dd t} ^{\ds{\Psi:\ Digamma\ Function}} \\[5mm] & = \left.\ln\pars{\Gamma\pars{t/6 + 2/3}\Gamma\pars{t/6 + 5/6} \over \Gamma\pars{t/6 + 1/6}\Gamma\pars{t/6 + 1/3}}\right\vert_{\ 0}^{\ 1}\qquad \pars{~\Gamma:\ Gamma\ Function.\ \Psi\pars{z} \stackrel{\mrm{def.}}{=} \totald{\ln\pars{\Gamma\pars{z}}}{z}~} \\[5mm] & = \ln\pars{{\Gamma\pars{5/6}\Gamma\pars{1} \over \Gamma\pars{1/3}\Gamma\pars{1/2}}\, {\Gamma\pars{1/6}\Gamma\pars{1/3} \over \Gamma\pars{2/3}\Gamma\pars{5/6}}} \\[5mm] & = \ln\pars{{1 \over \Gamma\pars{1/2}}\, {\Gamma\pars{1/6} \over \Gamma\pars{2/3}}}\qquad \pars{\ds{\mbox{Note that}\ \Gamma\pars{1} = 1\ \mbox{and}\ \Gamma\pars{1 \over 2} = \root{\pi}}} \\[5mm] & = \bbx{\ln\pars{{1 \over \root{\pi}}\, {\Gamma\pars{1/6} \over \Gamma\pars{2/3}}}} \approx 0.8412 \end{align}

Note that $\ds{\left.\vphantom{\Large A}\Psi\pars{z}\right\vert_{\ \Re\pars{z}\ >\ 0} = -\gamma + \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t}$. $\ds{\gamma}$ is the Euler-Mascheroni Constant. See $\ds{\color{#000}{\mathbf{6.3.22}}}$ in A & S Table.

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  • $\begingroup$ Many thanks for your nice and detailed answer. $\endgroup$ – user243301 Jul 17 '17 at 5:14
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    $\begingroup$ @user243301 You're welcome $\left)\substack{\bullet\quad\bullet\\ \mid\\ {\Huge\smile}}\right($. $\endgroup$ – Felix Marin Jul 17 '17 at 22:05
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I thought it might be instructive to present an approach that does not rely on Frullani's integral.

To that end, we first note that $\frac{x-1}{\log(x)}=\int_0^1 x^s\,ds$.

Therefore, we can write

$$\begin{align} \int_0^1 \frac{1+x}{1+x^3}\,\frac{x-1}{\log(x)}\,dx&=\int_0^1 \left(\int_0^1 \frac{x^s+x^{s+1}}{1+x^3} \right)\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty (-1)^n\int_0^1 (x^{s+3n}+x^{s+3n+1})\,dx\,ds\\\\ &=\int_0^1 \sum_{n=0}^\infty (-1)^n \left(\frac{1}{s+3n+1}+\frac{1}{s+3n+2}\right)\,ds\\\\ &=\sum_{n=0}^\infty (-1)^n \log\left(\frac{3n+3}{3n+1}\right)\\\\ &=\sum_{n=0}^\infty \log\left(\frac{(6n+3)(6n+4)}{(6n+1)(6n+6)}\right)\tag 1\\\\ &=\log\left(\frac{\Gamma(1/6)}{\sqrt{\pi}\Gamma(2/3)}\right)\tag2 \end{align}$$

where in going from $(1)$ to $(2)$ we relied on the analysis posted in Rob's solution herein.

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  • $\begingroup$ Many thanks for your nice answer, and approach. I am going to study it. $\endgroup$ – user243301 Jul 16 '17 at 19:55
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Jul 16 '17 at 20:10
  • $\begingroup$ The last = sign alone is a huge jump to anybody asking this, proposed with no justification. $\endgroup$ – Did Jul 17 '17 at 16:25
  • $\begingroup$ @Did There were already two posted solutions with a way forward on that huge jump. I thought it redundant to duplicate. Would you suggest that I duplicate those or just reference them? $\endgroup$ – Mark Viola Jul 17 '17 at 20:00
  • $\begingroup$ The sentence you added, at least makes clear that there is a gap. $\endgroup$ – Did Jul 17 '17 at 21:11

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