Evaluation of the definite Integral $\int_0^1 \sqrt [3] {x \log\left(\frac{1}{x} \right)}dx $

Question

What kind of substitution should i use and how can I get rid of $\log{\frac{1}{x}}$from the square root Note cube root is over both x and $\log{\frac{1}{x}}$.

Answer 1

Write the original problem as: $$I=\int_0^1 -x^{\frac{1}{3}}\ln^{\frac{1}{3}}(x)dx$$

Let: $~~\ln(x)=-u^3, ~~~x=e^{-u^3},~~~dx=-3u^2e^{-u^3} du$,

and plug in the substitution:

$$\begin{align} I=\int^\infty_0 3u^3\cdot e^{-\frac{4}{3}u^3}du \end{align}$$

Let: $~~t=\frac{4}{3}u^3$

$$\begin{align} I=\left(\frac{3}{4}\right)^{\frac{4}{3}}\int_0^\infty t^{\frac{1}{3}} e^{-t} dt=\left(\frac{3}{4}\right)^{\frac{4}{3}}\Gamma\left(\frac{4}{3}\right) \end{align}$$