How to calculate these gamma functions? Equation : $$\int _{0}^{\infty }x^{n}e^{-x}dx=n!=\Gamma(n+1) $$
1) $$ \int _{0}^{1}x^{2}\left( \ln \dfrac {1} {x}\right) ^{3}dx $$
2) $$\int _{0}^{1}\sqrt[3] {\ln x}dx $$
Hint : $$ x=e^{u} $$
3)Express as a Gamma Function.
$$ \int _{0}^{1}\left[ \ln \left( \dfrac {1} {x}\right) \right] ^{p-1}dx $$

$$\underbrace{\int_0^1 x^2 \ln^3 (1/x)dx = \int_{\infty}^{0} e^{-2t} \ln^3(e^t) \times -e^{-t} dt}_{x = e^{-t}} = \overbrace{\int_0^{\infty} t^3 e^{-3t} dt = \int_0^{\infty} \dfrac{z^3}{27} e^{-z} \dfrac{dz}3}^{t = z/3} = \dfrac{\Gamma(4)}{81}$$
$$\underbrace{\int_0^1 \sqrt[3]{\ln(x)}dx = \int_{\infty}^{0} \sqrt[3]{(-t)} \times - e^{-t} dt }_{x = e^{-t}} = - \int_0^{\infty} t^{1/3} e^{-t} dt = - \Gamma(4/3)$$
$$\underbrace{\int_0^1 \sqrt[p-1]{\ln(1/x)}dx = \int_{\infty}^{0} \sqrt[p-1]{t} \times - e^{-t} dt }_{x = e^{-t}} = \int_0^{\infty} t^{1/(p-1)} e^{-t} dt = \Gamma(p/(p-1))$$
