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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?

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

3 Answers 3

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$\int_0^{1}\frac 1 {1+x^{n}}dx =1-\int_0^{1}\frac {x^{n}} {1+x^{n}}dx$. 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 $1$.

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  • $\begingroup$ I accepted @Gae. S.'s answer and your answer is also wonderful. Sorry~ $\endgroup$
    – Syvshc
    Apr 20, 2020 at 14:43
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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: $$0\leq\frac{x^n}{1+x^{n}}\leq x^n\implies \lim_{n\to\infty}\int_{0}^{1}\frac{x^n}{1+x^{n}}\,dx=0.$$

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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 $$\liminf_{n\to\infty}\int_0^{1-n^{-1/2}}\frac1{1+x^n}\,dx\ge \frac1{1+0}(1-0)=1$$

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  • $\begingroup$ I understand how you handled it. I use this method on the $ 1-1/n $, and it also seems to work. $\endgroup$
    – Syvshc
    Apr 20, 2020 at 9:15
  • $\begingroup$ 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 $\endgroup$
    – user239203
    Apr 20, 2020 at 9:20
  • $\begingroup$ OH! I understand! thanks, I forgot that if I use $ 1-1/n $, I will lose the power of $ n^{1/2} $ $\endgroup$
    – Syvshc
    Apr 20, 2020 at 9:25
  • $\begingroup$ And should the last integral formula's upper limit be $ 1-n^{-1/2} $? $\endgroup$
    – Syvshc
    Apr 20, 2020 at 9:29
  • $\begingroup$ @Syvshc Ah, I see you were referring to the typo. $\endgroup$
    – user239203
    Apr 20, 2020 at 11:15

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