Compute the limit of $\int_{n}^{e^n} xe^{-x^{2016}} dx$ when $n\to\infty$

Question

Find the following limit $$I = \lim_{n \to\infty} \int_{n}^{e^n} xe^{-x^{2016}} dx$$

My attempt

Assumption: as $n \to \infty$ we can assume and interval on the positive real axis $[n,e^n]$

Here the function $e^{-x^{2016}}$ is a decreasing function, using this fact we use the sandwich lemma to evaluate I

$$LHS = e^{-(e^n)^{2016}} \int_{n}^{e^n}x dx \leq I \leq e^{-(n)^{2016}} \int_{n}^{e^n}x dx =RHS$$

Limit of $LHS$ and $RHS$ can be shown to be zero, hence $I=0$.

Evaluation of LHS $$\lim_{n \to \infty} e^{-(e^n)^{2016}} \frac{(e^n)^2 - n^2}{2} = \lim_{n \to \infty} e^{-(e^n)^{2016}} e^{2n}\frac{1 - \frac{n^2}{e^{2n}}}{2} = 0.$$

I need to know weather I can make as Assumption as I have made above. I also need some help in verifying weather the evaluation of limit LHS is done correctly.

Answer 1

Your solution is perfectly fine. But note that $f(x)=xe^{-x^{2016}}$ is decreasing monotonically for $x\ge \left(\frac{1}{2016}\right)^{1/2016}$. Hence, for $n\ge \frac{1}{2016}$, we can write

$$(e^n-n)e^ne^{-e^{2016\,n}}\le \int_n^{e^n}xe^{-x^{2016}}\,dx\le (e^n-n)e^{-n^{2016}}$$

whence applying the squeeze theorem yields the expected result.

Answer 2

The function $f(x) = x\cdot e^{-x^{2016}}\implies f'(x) = e^{-x^{2016}}+x\cdot e^{-x^{2016}}(-2016x^{2015})= e^{-x^{2016}}\left(1-2016x^{2016}\right)< 0$ on $[n,e^{n}]$. Thus you can find an upper bound and lower bound for the integral using $f_{\min}= f(e^{n}), f_{\max} = f(n)$, and the property that $f(e^n)\left(e^n-n\right)< I_n < f(n)\left(e^n-n\right)$. It is expected that both expressions have limits $0$, and "sandwich" lemma follows...