# find $\lim\limits_{n\to\infty}\int_{0}^{1}1/(1+x^{n})\,dx$

First of all, English is not my native language, but Chinses is. I tried to spilt the integration interval into 2 pieces: $$[0, 1-1/n]$$ and $$[1-1/n, 1]$$. In both intervals I use the mean value theorem: $$\int_{0}^{1-1/n}\frac{1}{1+x^{n}}\,dx=\frac{1}{1+\xi_{n}^{n}}\left( 1-\frac{1}{n} \right), \qquad \text{and} \qquad \int_{1-1/n}^{1}\frac{1}{1+x^{n}}\,dx=\frac{1}{1+\eta_{n}^{n}}\frac{1}{n},$$ where $$\xi_{n}\in(0, 1-1/n), \eta_{n}\in(1-1/n, 1)$$.I found that the latter formula has a limit of $$0$$ when $$n\to\infty$$. However I can't handle the previous formula. Does anyone has some thoughts?

• It seems it should be easy to use dominated convergence theorem. Also, why did you split the integration in two?
– lcv
Apr 20, 2020 at 8:50
• because the function sequence $\{1/(1+x^n)\}$ doesn't converge at $x=0$ Apr 20, 2020 at 8:55
• All of the answer is wonderful, and the answer which I accepted resumes my thought. Thank you all a lot :)@KaviRamaMurthy @Riemann Apr 20, 2020 at 14:46

$$\int_0^{1}\frac 1 {1+x^{n}}dx =1-\int_0^{1}\frac {x^{n}} {1+x^{n}}dx$$. Note that $$0 \leq \frac {x^{n}} {1+x^{n}} \leq x^{n}$$ and $$\int_0^{1} x^{n}dx=\frac 1 {n+1} \to 0$$. Put these together to see that the required limit is $$1$$.

• I accepted @Gae. S.'s answer and your answer is also wonderful. Sorry~ Apr 20, 2020 at 14:43

Consider limit $$\lim_{n\to\infty}\int_{0}^{1}\frac{x^n}{1+x^{n}}\,dx=0.$$ then your limit $$\lim_{n\to\infty}\int_{0}^{1}\frac{1}{1+x^{n}}\,dx=1.$$ Hint： $$0\leq\frac{x^n}{1+x^{n}}\leq x^n\implies \lim_{n\to\infty}\int_{0}^{1}\frac{x^n}{1+x^{n}}\,dx=0.$$

Consider $$\int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx$$ and $$\int_{1-n^{-1/2}}^1\frac1{1+x^n}\,dx$$ instead. The second integral is still bounded above by $$n^{-1/2}$$.

For the first integral, $$1\ge \int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx=\frac{1}{1+\xi_n^{n}}(1-n^{-1/2})\ge \frac1{1+(1-n^{-1/2})^n}(1-n^{-1/2})$$

However, $$\left(1-n^{-1/2}\right)^n=\left(\left(1-n^{-1/2}\right)^{n^{1/2}}\right)^{n^{1/2}}\stackrel{n\to\infty}{\longrightarrow} \left[\left(e^{-1}\right)^{\infty}\right]=0$$, therefore $$\liminf_{n\to\infty}\int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx\ge \frac1{1+0}(1-0)=1$$

• I understand how you handled it. I use this method on the $1-1/n$, and it also seems to work. Apr 20, 2020 at 9:15
• I guess you could still say that $\int_0^{1-n^{-1}}\ge \int_0^{1-n^{-1/2}}$ and go from there with my estimate, but otherwise I don't see you going around the fact that $\lim_{n\to\infty}(1-n^{-1})^n=e^{-1}$, thus giving the loose lower bound $\frac1{1+e^{-1}}$ for the integral. @Syvshc
– user239203
Apr 20, 2020 at 9:20
• OH! I understand! thanks, I forgot that if I use $1-1/n$, I will lose the power of $n^{1/2}$ Apr 20, 2020 at 9:25
• And should the last integral formula's upper limit be $1-n^{-1/2}$? Apr 20, 2020 at 9:29
• @Syvshc Ah, I see you were referring to the typo.
– user239203
Apr 20, 2020 at 11:15