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Prove $$\int_0^{\infty} \frac{\ln^2{(x^3+1)}}{x^3+1} \; \mathrm{d}x = \frac{\sqrt{3} \pi}{18} \left(9\ln^2{(3)}+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\frac{\pi^3 \sqrt{3}}{54}-\frac{\pi}{3}\ln{(3)}$$ I tried feynman method with $ I(a)=\int_0^{\infty} \frac{\ln^2\left(ax^3+1\right)}{x^3+1} \; \mathrm{d}x$ but this got ugly because we need to differentiate wrt a twice I think. I also try to factor $x^3+1$ and either partial fraction decomposition denominator or log property for numerator but these did not work. I am not sure what to do now. Im not very good with contour integration so can people who respond try to use real methods?? Any help is appreciated

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Consider the parameterized integral $I(a)$ where the integral in question is equal to $I''(1)$:

$$I(a)=\int_0^{\infty} \frac{1}{\left(x^3+1\right)^a} \; \mathrm{d}x$$ First, let $x^3 \to x$: \begin{align} I(a) &= \frac{1}{3} \int_0^{\infty} \frac{t^{-\frac{2}{3}}}{(t+1)^a} \notag \\ & \; \; = \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)}{\Gamma\left(a\right)} \notag \end{align} Where we used the definition of the beta function.

Now we will find $I''(1)$: \begin{align} I'(a) &= \frac{d}{da} \left( \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)}{\Gamma\left(a\right)} \right) \notag \\ &= \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)\left(\psi\left(a-\frac{1}{3}\right)-\psi\left(a\right)\right)}{\Gamma\left(a\right)} \notag \\ I''(a) &= \frac{d}{da} \left( \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)\left(\psi\left(a-\frac{1}{3}\right)-\psi\left(a\right)\right)}{\Gamma\left(a\right)} \right) \notag \\ &= \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)\left(\psi\left(a-\frac{1}{3}\right)^2-2\psi\left(a\right)\psi\left(a-\frac{1}{3}\right)+\psi\left(a\right)^2+\psi^{\prime}\left(a-\frac{1}{3}\right)-\psi^{\prime}\left(a\right)\right)}{\Gamma\left(a\right)} \notag \\ I''(1) &= \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(\frac{2}{3}\right)\left(\psi\left(\frac{2}{3}\right)^2-2\psi\left(1\right)\psi\left(\frac{2}{3}\right)+\psi\left(1\right)^2+\psi^{\prime}\left(\frac{2}{3}\right)-\psi^{\prime}\left(1\right)\right)}{\Gamma\left(1\right)} \notag \\ &= \frac{2\pi}{3\sqrt{3}}\left(-\frac{\pi^2}{12}+\frac{9\ln^2{(3)}}{4}-\frac{\pi \sqrt{3} }{2} \ln{(3)} + \psi^{\prime} \left(\frac{2}{3}\right)\right) \notag \\ &= \boxed{\frac{\sqrt{3} \pi}{18} \left(9\ln^2{(3)}+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\frac{\pi^3 \sqrt{3}}{54}-\color{red}{\frac{\pi^2}{3}\ln{(3)}}} \notag \end{align} It appears that the result of the integral you provided is slightly off (see the red term). Wolfram Alpha agrees with the answer I provided as well.

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  • $\begingroup$ Maybe answer key had a typo. Thanks $\endgroup$ – user801111 Aug 12 '20 at 15:25
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    $\begingroup$ The same 5 down votes hours after a perfectly correct accepted answer. Someone has some sort of a vendetta and just uses 5 accounts to downvote my answers or what? This is at least the third time this has happened within a couple of days. I assume its the same 5 people who upvoted Dennis Orton's comment since it all happened at around the same time. $\endgroup$ – Ty. Aug 13 '20 at 0:28
  • $\begingroup$ This looks like a serious issue. Perhaps you should flag this answer and tell moderators what happened? $\endgroup$ – pisco Aug 13 '20 at 17:13
  • $\begingroup$ @pisco. I flagged the answer. Thanks for the idea. It is just strange how this started recently and occurs on some but not all of my answers. $\endgroup$ – Ty. Aug 13 '20 at 17:18
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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} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x^{3} + 1} \over x^{3} + 1}\,\dd x} \,\,\,\stackrel{\large x^{3}\ \mapsto\ x}{=}\,\,\, \int_{0}^{\infty}{\ln^{2}\pars{x + 1} \over x + 1}\, {1 \over 3}\,x^{-2/3}\,\dd x \\[5mm] = &\ \left.{1 \over 3}\,\partiald[2]{}{\nu}\int_{0}^{\infty}x^{\color{red}{1/3} - 1}\,\pars{x + 1}^{\nu - 1}\,\dd x\,\right\vert_{\ \nu\ =\ 0}\label{1}\tag{1} \end{align} \begin{align} \mbox{However,}\quad \pars{x + 1}^{\nu - 1} & = \sum_{k = 0}^{\infty} {\nu - 1 \choose k}x^{k} = \sum_{k = 0}^{\infty}\bracks{{k - \nu \choose k}\pars{-1}^{k}}x^{k} \\[5mm] &= \sum_{k = 0}^{\infty} \color{red}{\Gamma\pars{k - \nu + 1} \over \Gamma\pars{1 - \nu}}\, {\pars{-x}^{k} \over k!} \end{align} With the Ramanujan Master Theorem, (\ref{1}) becomes: \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x^{3} + 1} \over x^{3} + 1}\,\dd x} = \left.{1 \over 3}\,\partiald[2]{}{\nu} \bracks{\Gamma\pars{1 \over 3}{\Gamma\pars{-1/3 - \nu + 1} \over \Gamma\pars{1 - \nu}}} \,\right\vert_{\ \nu\ =\ 0} \\[5mm] = &\ \bbx{{2 \over 9}\,\root{3}\pi\,\Psi\, '\pars{2 \over 3} + {1 \over 2}\,\root{3}\pi\ln^{2}\pars{3} - {1 \over 3}\pi^{2}\ln\pars{3} - {1 \over 54}\,\root{3}\pi^{3}} \\[5mm] \approx &\ 2.3798 \end{align}

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