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.

$$\int _{0}^{1}\left[ \ln \left( \dfrac {1} {x}\right) \right] ^{p-1}dx$$

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welcome to math@SE. Please tell us where the problem is from (homework perhaps?) and what you have tried so far. – nbubis Jan 12 '13 at 20:22
@Erbil As the hint says, try plugging in $x=e^u$. You will be able to immediately recognize it in terms of $\Gamma$ function. – user17762 Jan 12 '13 at 20:28
I removed the tag functional-analysis, and added integral, gamma-function instead. – Eckhard Jan 12 '13 at 20:29
I think the hint pretty much gives it away. Erbil, you should explain what happened when you tried that substitution. – Michael Hardy Jan 12 '13 at 20:29
M. L. Boas-Mathematical Methods in the Physical Sciences - John Wiley (Chapter 11,Problems Sections 3) Here is the all questions. physicsforums.com/… 2) I have found it from table. 3) I have used recursion and table to find it. 4) Again With recursion. 5)Γ(0.7)= 1/p(p+1) with this formula. 8) IfGamma(p+1) is equal to this integral,I think it can be written as Gamma(2/3+1) 11) $$\int _{0}^{\infty }x^{2}e^{-x^{2}}dx=2\int _{0}^{\infty }u^{2}e^{-u}du=2\cdot \Gamma 3$$ NO idea for these questions. – Erbil Jan 12 '13 at 20:30

$$\underbrace{\int_0^1 x^2 \ln^3 (1/x)dx = \int_{\infty}^{0} e^{-2t} \ln^3(e^t) \times -e^{-t} dt}_{x = e^{-t}} = \overbrace{\int_0^{\infty} t^3 e^{-3t} dt = \int_0^{\infty} \dfrac{z^3}{27} e^{-z} \dfrac{dz}3}^{t = z/3} = \dfrac{\Gamma(4)}{81}$$
$$\underbrace{\int_0^1 \sqrt[3]{\ln(x)}dx = \int_{\infty}^{0} \sqrt[3]{(-t)} \times - e^{-t} dt }_{x = e^{-t}} = - \int_0^{\infty} t^{1/3} e^{-t} dt = - \Gamma(4/3)$$
$$\underbrace{\int_0^1 \sqrt[p-1]{\ln(1/x)}dx = \int_{\infty}^{0} \sqrt[p-1]{t} \times - e^{-t} dt }_{x = e^{-t}} = \int_0^{\infty} t^{1/(p-1)} e^{-t} dt = \Gamma(p/(p-1))$$
@Erbil $\ln(e^t) = t$. Hence, $$\ln^m(e^t) = (\ln(e^t))^m = t^m$$ – user17762 Jan 12 '13 at 21:25
@Erbil Note that under the transformation $x=e^{-t}$, for $x=0$ we need $e^{-t} = 0 \implies t = \infty$. Similarly, under the transformation $x=e^{-t}$, for $x=1$ we need $e^{-t} = 1 \implies t = 0$. – user17762 Jan 12 '13 at 21:55