$\lim_{n\to\infty}(\int_0^1 \frac{1}{1+x^n} dx)^n$

Question

$$\lim_{n\to\infty}\left(\int_0^1 \frac{1}{1+x^n} dx\right)^n$$

I solved this problem as in the link: Click here I got the answer as $0$ but it says incorrect. I think the answer is wrong or there is some error with the question.

My try: $$=\lim_{n\to\infty}\left(\int_0^1 \sum_{m=0}^{\infty} (-1)^mx^{mn}dx\right)^n$$ $$=\lim_{n\to\infty}\left( \sum_{m=0}^{\infty} \frac{(-1)^m}{1+mn}\right)^n$$ $$=\color{blue}{\lim_{n\to\infty}\left( \frac{\Phi(-1,1,\frac1n)}{n}\right)^n=0 ????}$$

Do you think the problem is incorrect or answer is incorrectly provided by the author ?

Answer 1

You may just use

• $\lim_{t\to 0}(1-t)^{\frac 1t} = e^{-1}$

Now, write

\begin{eqnarray*}\left(\int_0^1 \frac{1}{1+x^n} dx\right)^n & = & \left(1- \underbrace{\int_0^1 \frac{x^n}{1+x^n} dx}_{=:J_n}\right)^n \end{eqnarray*}

Using integration by parts you get $$J_n = \frac 1n\int_0^1 x\frac{nx^{n-1}}{1+x^n} dx = \frac 1n\left(\ln 2 - \underbrace{\int_0^1 \ln (1+x^n) dx}_{=:I_n}\right)$$

Now, DCT (or just squeezing) gives immediately $$\lim_{n\to \infty }I_n = 0$$

All together

\begin{eqnarray*}\left(\int_0^1 \frac{1}{1+x^n} dx\right)^n & = & \left(\left(1-\frac{\ln 2 - I_n}n\right)^{\frac{n}{\ln 2 - I_n}}\right)^{\ln 2 - I_n} \\ & \stackrel{n\to\infty}{\longrightarrow} & \left(e^{-1}\right)^{\ln2} = \frac 12 \end{eqnarray*}

Answer 2

Your answer cannot be correct.

Since, by your argument, $$\int_{0}^{1} \frac{dx}{1+x^n}>1-\frac{1}{1+n}$$

A lower bound for your limit is:

$$\left(1-\frac1{n+1}\right)^n\to \frac1e$$

Answer 3

Let $I_n$ be given by the integral $I_n=\int_0^1 \frac1{1+x^n}\,dx$. Enforcing the substitution $x\mapsto x^{1/n}$ and reveals

$$\begin{align} I_n&=\frac1n \int_0^1 \frac{x^{1/n}}{x(1+x)}\,dx\tag1 \end{align}$$

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Using partial fraction expansion on the integrand in $(1)$, we obtain

$$\begin{align} I_n&=\frac1n \int_0^1 x^{1/n-1}\,dx-\frac1n \int_0^1 \frac{x^{1/n}}{1+x}\,dx\\\\ &=1-\frac1n \int_0^1 \frac{x^{1/n}}{1+x}\,dx\tag2 \end{align}$$

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Next, using $1+x\le e^x\le \frac1{1-x}$ for $x<1$ provides the estimates

$$e^{\frac1n\log(x)}-1\ge \frac{\log(x)}{n}$$

and for $x\in (0,1)$

$$\begin{align} e^{\frac1n\log(x)}-1\le \frac{\frac{\log(x)}{n} }{1-\frac{\log(x)}{n}}\le \frac{\log(x)}{n}+\left(\frac{\log(x)}{n} \right)^2 \end{align}$$

Using these estimates in $(2)$, we can assert that

$$I_n=1-\frac{\log(2)}{n}+O\left(\frac1{n^2}\right)\tag3$$

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Finally, raising both sides of $(3)$ to the $n$'th power and letting $n\to \infty$ yields the coveted limit

$$\begin{align} \lim_{n\to \infty}I_n^n&=\lim_{n\to \infty}\left(1-\frac{\log(2)}n+O\left(\frac1{n^2}\right)\right)^n\\\\ &=\frac12 \end{align}$$

Answer 4

Picture of MMA input as requested: