# Prove $\int_0^{\infty} \frac{\ln^2(x^3+1)}{x^3+1} dx = \frac{\sqrt{3} \pi}{18} \left(9\ln^2(3)+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\ldots$

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

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.
$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \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}