$\displaystyle I_1 = \int^1_0 \sqrt[3]{\log( x)}\,dx$
I am trying to write this in terms of gamma functions.
After I substitute
$\displaystyle y = \log( x)$ , $x = e^y$ , $dx = e^y dy$
Then,
$\displaystyle I_1 = \int^1_0 e^yy^{\frac{1}{2}}\,dy$
but the upper limit isn't infinite. Can anyone please explain or help me?
The limits are not correct. With $y=\log(x)$, note that when $x\to 0^+$, $y\to -\infty$ and when $x=1$, $y=0$. With that substitution we find that
$$\int_0^1 \log^{1/3}(x)\,dx=\int_{-\infty}^0 y^{1/3}e^y\,dy$$
A slightly more direct substitution is to let $x=e^{-y}$ so that as $x\to 0^+$, $y\to \infty$, and for $x=1$, $y=0$, and $dx=-e^{-y}\,dy$.
Then, we can write
$$\int_0^1 \log^{1/3}(x)\,dx=-\int_0^\infty (y)^{1/3}e^{-y}\,dy=-\Gamma(4/3)$$