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

Question

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

Answer 1

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.