Reverse engineer the gamma question from a typo $\int_0^1(\ln1/x)^{1/3}$dx and the answer $\Gamma(4/3)$

Question

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: $$\int_0^1(\ln(x)/x)^{1/3}\mathrm{dx}$$ $$x=e^{-u}$$ $$-\int_0^\infty (-u)^{1/3}*e^{-2u/3}(-\mathrm{du})$$ $$\int_0^\infty (-3/2u)^{1/3}*e^{-u}*(\frac{3}{2}\mathrm{du})$$

Answer 1

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.

Answer 2

Why don't you write down the whole process to find out mistakes? The following is what I get:

$$I:=\int_0^1\left(\frac{\log x}x\right)^{1/3}dx\;:\;\;\text{substitution}\;\;e^{-u}:=x\implies-e^{-u}du=dx\implies$$

$$I=\int_\infty^0\left(\frac{(-u)}{e^{-u}}\right)^{1/3}(-e^{-u}du)=-\int_0^\infty u^{1/3}e^{-2u/3}du$$

and now a new substitution:

$$t=\frac{2u}3\implies du=\frac32dt\implies I=-\sqrt[3]\frac32\int_0^\infty t^{1/3}e^{-t}\frac32dt=-\sqrt[3]{\left(\frac32\right)^4}\,\Gamma\left(\frac43\right)$$

Answer 3

Starting with the gamma function $\int_0^\infty x^{1/3}e^{-x}$dx, and making the reverse substitution ln(1/x)=u (found by solving the original substitution $e^{-u}$) yields the intended question: $$\int_0^1(\ln(1/x))^{1/3}\mathrm{dx}$$ with solution $\Gamma(4/3)$