How to solve $\int_0^1\:\frac{\ln(x)\:\Big[1+x^{-\frac{1}{3}}\Big]}{(1-x)\sqrt[3]{x}}\:dx$

Question

How can we solve this integral? $\int_0^1\:\frac{\ln(x)\:\Big[1+x^{-\frac{1}{3}}\Big]}{(1-x)\sqrt[3]{x}}\:dx$

Answer 1

Subbing $x=u^3$ is very natural here and leads to a simple sum...

$$I = \int_0^1 dx \frac{\left (1+x^{-1/3} \right ) \log{x}}{x^{1/3} (1-x)} = 9 \int_0^1 du \, \frac{(1+u) \log{u}}{1-u^3}$$

Expand the denominator...

$$\begin{align}I &= 9 \sum_{k=0}^{\infty} \int_0^1 du \, (1+u) u^{3 k} \log{u}\\ &= -9 \sum_{k=0}^{\infty} \left (\frac1{(3 k+1)^2} + \frac1{(3 k+2)^2} \right ) \\ &= -9 \frac{\pi^2}{6} + 9 \sum_{k=0}^{\infty} \frac1{(3 k+3)^2}\\ &= - \frac{9\pi^2}{6} + \frac{\pi^2}{6}\\ &= -\frac{4 \pi^2}{3}\end{align}$$

Answer 2

We have

$$\int_0^1 x^a dx = \frac{1}{a+1}$$

$$\int_0^1 x^a \log x ~dx = -\frac{1}{(a+1)^2}$$

$$\int_0^1 x^b x^q \log x ~dx = -\frac{1}{(b+q+1)^2}$$

$$\int_0^1 \sum_{b=0}^{\infty} x^b x^q \log x ~dx = -\sum_{b=0}^{\infty} \frac{1}{(b+q+1)^2}$$

$$\int_0^1 \frac{x^q \log x}{1-x} ~dx = -\sum_{b=0}^{\infty} \frac{1}{(b+q+1)^2}=-\zeta(2,q+1)=-\psi^{(1)}(q+1) $$

With the last two expressions being Hurwitz zeta and polygamma functions respectively.

So, in your case:

$$\int_0^1 \frac{(x^{-1/3}+x^{-2/3}) \log x}{1-x} ~dx =-\zeta \left(2,\frac{2}{3} \right)-\zeta \left(2,\frac{1}{3} \right) =-\psi^{(1)} \left(\frac{2}{3} \right)-\psi^{(1)} \left(\frac{1}{3} \right)$$

Using the reference provided by @tired

$$\psi^{(1)} \left(z \right)+\psi^{(1)} \left(1-z \right)=\frac{\pi^2}{\sin^2 (\pi z)}$$

$$\int_0^1 \frac{(x^{-1/3}+x^{-2/3}) \log x}{1-x} ~dx =-\frac{4\pi^2}{3}$$