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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!

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1 Answer

up vote 4 down vote accepted

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

$$\int\limits_0^{ + \infty } {{e^{ - u}}{u^\lambda }du} \sim {\left( {\frac{\lambda }{e}} \right)^\lambda }\sqrt {2\pi \lambda } $$

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Thank you Peter! That substitution made it a lot easier! –  Maria May 30 '12 at 2:45
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