# How to find the limit $\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx$

Find the limit $$\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx$$

By intuition, I can guess the answer is $$0$$, but I have no idea how to start to prove it.

What I have tried is using the mean value theorem. There exists $$c \in (0, 1)$$: $$\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx = \lim_{n \to \infty}\frac{c^ne^c}{1 +e^c}$$ and I don't see how I should continue from here.

• Try the dominant convergence theorem. – CO2 Dec 15 '20 at 14:50
• Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts. – GNUSupporter 8964民主女神 地下教會 Dec 15 '20 at 14:51
• @BiAo the $e^c/(1+e^c)$ is a constant and since $0<c<1$ then $c^n\to 0$ as $n\to\infty$. – Pixel Dec 15 '20 at 15:09
• @Pixel The $c$ will tipically depend on $n$. What if $c$ approaches $1$? For example $c \sim 1-\frac{1}{n}$? – Gary Dec 16 '20 at 6:49
• @Gary yes that does seem problematic. I tried some substitutions but the carpet still doesn't fit. A $1-\varepsilon$ argument may work as DonAntonio mentioned, but not sure. – Pixel Dec 16 '20 at 13:21

From $$0\le \int_0^1{\frac{x^ne^x}{1+e^x}}dx=\int_0^1{\frac{x^n}{1+e^{-x}}}dx\le \int_0^1{\frac{x^n}{1}}dx=\int_0^1{x^n}dx$$ From the squeeze theorem, take limits on both sides and we get $$0\le \lim_{n\rightarrow \infty} \int_0^1{\frac{x^ne^x}{1+e^x}}dx=\lim_{n\rightarrow \infty} \int_0^1{\frac{x^n}{1+e^{-x}}}dx\le \lim_{n\rightarrow \infty} \int_0^1{x^n}dx=0$$ and thus $$\lim_{n \to \infty} \int_0^1\frac{x^ne^x}{1 +e^x}dx=0$$
We have $$0 < \frac{{e^x }}{{1 + e^x }} < 1$$ for $$0, thus $$0 < \int_0^1 {\frac{{x^n e^x }}{{1 + e^x }}dx} < \int_0^1 {x^n dx} = \frac{1}{{n + 1}}.$$ Consequently, by the squeeze theorem, $$\mathop {\lim }\limits_{n \to + \infty } \int_0^1 {\frac{{x^n e^x }}{{1 + e^x }}dx} = 0.$$
Since the integrand is a continuous function, the mean value theorem for integrals tells us that there exists $$\;0 s.t.
$$\int_0^1x^n\frac{e^x}{1+e^x}\,dx=c^n\frac{e^c}{1+e^c}\xrightarrow[n\to\infty]{}0$$
since $$\;c^n\to 0\;$$ and $$\;0\le \cfrac{e^x}{1+e^x}\le 1\;$$ is bounded.
• The $c$ will tipically depend on $n$. What if $c$ approaches $1$? For example $c \sim 1-\frac{1}{n}$? – Gary Dec 16 '20 at 6:47
• @Gary Very good point, though taking the upper limit as $\;1-\epsilon\;$ could solve the problem...perhaps. I shall take a second thought on this one. Observe that $\;c_n\to 1\;$ is the only problem here. Any other behaviour, like swinging away from $\;1\;$ or converging to some $\;h\in [0,1)\;$ would yield the same result as above. – DonAntonio Dec 16 '20 at 8:49