Reflect on Euler

First observe that digamma function is

$\displaystyle \psi_0(n+1) = \int_0^1{\frac{1-x^n}{1-x} \ dx}$

Now, take 2 derivatives,

$\displaystyle \psi_2(n+1) = \int_0^1{\frac{\ln^2(x) x^n}{x - 1} \, dx}$

Now, our problem is

$\displaystyle \int_0^1{\ln^2(x) \frac{x^{-1/3} - x^{-2/3}}{x-1} \, dx}$

Using our above result,

$\displaystyle = \psi_2\left(1-\frac{1}{3}\right) - \psi_2\left(1-\frac{1}{3}\right)$

$\displaystyle = \psi_2 \left(\frac{2}{3}\right) - \psi_2\left(\frac{1}{3}\right)$

Now, using Reflection formula $\displaystyle \psi_n(1-z) + (-1)^{n+1}\psi_n(z) = (-1)^n \pi \dfrac{d^n}{dz^n}\left(\cot(\pi z)\right)$ ,

$\displaystyle = (-1)^2 \pi \dfrac{d^2}{dz^2}\left(\cot(\pi z)\right) = \pi \left(2\pi^2 \cot(\pi z) \csc^2(\pi z)\right)|_{z=1/3}$

$\displaystyle = \dfrac{8\pi^3}{3\sqrt{3}}$