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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.

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    If you showed that LHS and RHS are zero, you have done! – 2017-02-27
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    I am not very sure abt the limit evaluation hence i need help there, is the comparision and the assumption that i can consider an interval on the positive side of the real axis correct? – 2017-02-27
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    $x^{2016}>n^{2016}>n^2$ so $\exp(-x^{2016})<\exp(-n^2)$ – 2017-02-27
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    Isn't the function inside the integral integrable on the entire real line? – 2017-03-20
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    @MarkoKarbevski yes indeed – 2017-03-20
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    In that case, the $exp(n)$ is redundant and only there to confuse you. Let $f$ be integrable on the positive reals. What is the limit of $\int_{[t,\infty)} f(x)dx$ as $t$ goes to infinity? – 2017-03-20
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    @MarkoKarbevski could you explain a bit further i get the intuition ...but how would you put it rigorously – 2017-03-20
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    @spaceman_spiff You can, for example, use Lebesgue's dominated convergence theorem. – 2017-03-20
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    ok thanks but i havent covered them...i am an undergrad..this question is from a masters entrance exam ...so i think i am expected to use squeeze thm here... – 2017-03-20
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    @spaceman_spiff As long as the solution is correct you are fine. But I still think that it is more clarifying to use the machinery of integral convergence. It can be proved using elementary methods too (I would even suggest trying it, one way to prove it is by contradiction). In terms of series, the similar result is that if $\sum_{n>0} a_n$ converges then $\sum_{n>k} a_n$ converges to 0 as $k \to \infty$ which can be done by using the Cauchy criterion for converging sequences. – 2017-03-20

2 Answers 2

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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.

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    kindly note that f(x) used by you is not what i have claimed to be decreasing on the positive x axis. – 2017-02-27
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    @spaceman_spiff Yes, your solution is fine. I am offering another way forward that leads to the same result as yours. – 2017-02-27
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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...