Reverse engineer the gamma question from a typo $\int_0^1(\ln1/x)^{1/3}$dx and the answer $\Gamma(4/3)$ Background: question 9.1.2.d in a newly published book Mathematics for Physical Science by Harris.
My question is what was the question supposed to be? Taking ln(x) for ln1 I'm doing something wrong or not guessing the right typo:
$$\int_0^1(\ln(x)/x)^{1/3}\mathrm{dx}$$
$$x=e^{-u}$$
$$-\int_0^\infty (-u)^{1/3}*e^{-2u/3}(-\mathrm{du})$$
$$\int_0^\infty (-3/2u)^{1/3}*e^{-u}*(\frac{3}{2}\mathrm{du})$$

Notice that:
$$\int_0^\infty \frac{3}{2}\left(-3/2u\right)^{\frac{1}{3}}e^{-u}\mathrm{du} = \int_0^\infty \frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}u^{\frac{1}{3}}e^{-u}\mathrm{du} = \\
\frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\int_0^\infty u^{\frac{4}{3}-1}e^{-u}\mathrm{du} =\frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\Gamma\left(\frac{4}{3}\right). $$
Recall also that $$\Gamma(z) = (z-1)\Gamma(z-1).$$
Then:
$$\frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\Gamma\left(\frac{4}{3}\right) = \frac{3}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right) =  \frac{1}{2}\left(-\frac{3}{2}\right)^{\frac{1}{3}}\Gamma\left(\frac{1}{3}\right).$$
The results is confirmed also by Wolframalpha. Actually, you can't get rid of the multiplied term in front of the final expression.
