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!
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 } $