Estimate a value of $\int_2^\infty \frac{x^k e^{-x}}{x \left( \log(x) \right)^2}\, dx $

Question

How to estimate a value of the following integral \begin{align} \int_2^\infty \frac{x^k e^{-x}}{x \left( \log(x) \right)^2} \,dx, \end{align} for $k=0,1, \dots$.

Answer 1

This solution is guided by concepts from the asymptotic estimation of integrals. For $k$ large, the integrand looks like a gaussian with its center offset much further than its width from the origin. Find the stationary point of argument of the exponential, $\exp((k-1)\log{x} -x)),$ which is $x_0=k-1.$ Since the logarithm is slowly varying, evaluate it at $x_0$ and pull it through the integral. You are then left with the integrand of the gamma function. There is no reason why the lower integral limit can't be extended to 0 because by assumption $k$ is large enough that you are on the far left side of the peak, which is nearly zero for $x<2.$ Thus you get,

$$ \int_2^\infty e^{-x}\,x^{k-1}\frac{dx}{\log^2{x}} \sim \frac{\Gamma(k)}{\log^2(k-1)}. $$

The approximation is within 1% for $k$ as small as 13. Use the Stirling approximation for the gamma function if you need a more elementary form.