Finding the asymptotic limit of an integral. I'm having trouble finding the asymptotic of the integral
$$ \int^{1}_{0} \ln^\lambda \frac{1}{x} dx$$
as $\lambda \rightarrow + \infty$.
Can anyone help? 
Thank you! 

Let 
$-\log x=u$ then the integral becomes
$$\int\limits_0^1 {{{\left( { - \log x} \right)}^\lambda }dx}  = \int\limits_0^{ + \infty } {{e^{ - u}}{u^\lambda }du} $$
This is Euler's famous Gamma function, which has an asymptotic formula by Stirling
$$\int\limits_0^{ + \infty } {{e^{ - u}}{u^\lambda }du}  \sim {\left( {\frac{\lambda }{e}} \right)^\lambda }\sqrt {2\pi \lambda } $$
