I am trying to show the above using the Dominated convergence theorem. I have let the integrand be the limit point of a sequence of functions obtained by replacing the log by its expansion ($\log(\frac{1}{x})=\log((\frac{1}{x}-1)+1)$). I have reduced the problem to finding the integral:
$$\int_{0}^{1}{x^{\frac{1}{3}-k}(1-x)^{k-1}}dx$$ but I feel Cike this is the wrong approach. How can this be shown using The Dominated convergence Theorem, and more specifically how can we choose an appropriate sequence of functions?
Let $x = e^{-t} \therefore dx=-e^{-t}dt$. The integral is, $$I=\int_{0}^{\infty}\frac{te^{-t/3}}{1-e^{-t}}dt$$ $\because 0 < e^{-t} <1$ for $t \in (0,\infty)$, we can express the denominator as a geometric progression. $$I=\int_{0}^{\infty}te^{-t/3}(1+e^{-t}+\cdots)dt$$ $\because \int_0^{\infty}te^{-at}dt=\frac{1}{a^2}$, we get, $$I=3^2+\frac{3^2}{4^2}+\frac{3^2}{7^2}\cdots=9\sum_{n=1}^{\infty}\frac{1}{(3n+1)^2}$$ which is the desired result. Hope this helps.