I am looking for an equivalent (or something weaker) of the following integral:
$$u_n=\int_{1}^{+\infty}e^{-t}\log^n(t)\,\mathrm dt$$
when $n\rightarrow\infty$. I have tried some recurrence relations that leads to $u_n^{\frac{1}{n}}=o(n^{\epsilon})$ for any $\epsilon>0$, but I'm sure it is possible to get something sharper. Any hint is welcome! Thank you.
If a heuristic consideration is acceptable, we can do the following. $$I(n)=\int_1^\infty e^{-t}\ln^nt\,dt\overset{t=nx}{=}n\int_{1/n}^\infty e^{-n\big(x-\ln(\ln n+\ln x)\big)}dx=n\int_{1/n}^\infty e^{-nf(x,n)}dx$$ Though the function $f(x,n)=x-\ln(\ln n+\ln x)$ depends on $n$, we can try to apply the Laplace method. Logarithm is a very slowly changing function, and the extremum of $f(x,n)$ lies nearly $x=\frac1{\ln n};$ (can be easily checked by iterations).
For $n\gg1$ the point $x=\frac1{\ln n}$ lies far from the end of the interval of integration ($x=\frac1n$), so we may suppose that the Laplace' method is applicable (checking afterwards that we get a reasonable approximation). $$\frac{d}{dx}f(x,n)=1-\frac1{x\ln(xn)}=0\,\,\Rightarrow\,\,x\ln(xn)=1\,\,\Rightarrow\,\,\frac1xe^\frac1x=n$$ so the extremum (maximum) point is $x_0=\frac1{W(n)}$, where $W$ is the Lambert's function, defined as $\,We^W=n$ $$f''(x,n)=\frac{1+\ln(xn)}{x^2\ln^2(xn)}; \,f(x_0;n)=W(n)+1$$ $$I(n)\approx ne^{-n\left(\frac1{W(n)}-\ln W(n)\right)}\int_{1/n}^\infty e^{-\frac n2\big(W(n)+1\big)\left(x-\frac1{W(n)}\right)^2}dx$$ $$\approx ne^{-n\left(\frac1{W(n)}-\ln W(n)\right)}\int_{-\infty}^\infty e^{-\frac n2\big(W(n)+1\big)t^2}dt$$ $$\boxed{\,\,I(n)\approx\sqrt{\frac{2\pi n}{W(n)+1}}e^{n\left(\ln W(n)-\frac1{W(n)}\right)}=\sqrt{\frac{2\pi n}{W(n)+1}}(W(n))^ne^{-e^{W(n)}}\,\,}$$ For the rough approximation we can take $W(n)\approx \ln n$ (for example, for $W(n)=5,\, n=742.066$, and $\ln (742.066)=6.609...$), though to get higher accuracy we should use the exact value of $W(n)$ for every $n$ (or vice versa - evaluate $n$ for every $W$).
Numeric check with WolframAlpha
