Prove that $\lim\limits_{n\rightarrow\infty}\int_{1}^{a}\frac{1}{1+x^{n}}dx=\ln2$

Question

Prove that $$\lim_{n\rightarrow\infty}\int_{1}^{a}\frac{1}{1+x^{n}}\,\mathrm{d}x=\ln2.$$ We can write $$\frac{1}{1+x^{n}}=1-\frac{x^{n}}{1+x^{n}},$$ so $$\int_{1}^{a}\frac{1}{1+x^{n}}\,\mathrm{d}x=a-1-\int_{1}^{a}\frac{x^n}{1+x^n}\,\mathrm{d}x$$ and $$\int_{1}^{a}\frac{x^n}{1+x^n}\,\mathrm{d}x=\frac{1}{n}x\ln(1+x^n)|_{1}^{a}+\int_{1}^{a}\ln(1+x^n)\,\mathrm{d}x$$

The last can be manipulated to providing a limit using $e^x\geq x+1$, in our case $x \geq \ln(x+1)$, $x \geq0$. However, I need help from this point on or, if the method is faulty, on the problem itself. EDIT: Thanks for clarifications. I see the mistake. However, by changing it into$\lim_{n\rightarrow\infty}n\int_{1}^{a}\frac{1}{1+x^{n}}\,\mathrm{d}x=\ln2.$ I followed Cosmin's duplicate, but I am unable to grasp the proof, as the level is high. Could you give an easier proof for this than one using improper integrals, please?

Answer 1

This cannot be true because of Prove that $\lim n\int_1^a\frac{1}{1+x^n}dx=\ln 2$ .

If we would have that $$\int_1^a \frac{1}{x^n+1}dx \to \ln (2) \text{ as } n \to \infty,$$ then we would get that $$n \int_1^a \frac{1}{x^n+1}dx \to \infty \text{ as } n \to \infty,$$ which is false.

Unfortunately, for the sequence of functions $(f_n)$ defined as $$f_n: [1,a] \to \mathbb{R}, f_n(x) = \frac{1}{1+x^n}, \forall n \in \mathbb{Z}, n \geq 0,$$ we do not have uniform convergence to $f \equiv 0$ on $[1,a],$ but we do have uniform convergence on $(1,a]$ to $f$. Then you can compute $$\lim_{n \to \infty} \left( \int_{1+\epsilon}^a f_n(x)dx \right), $$ for every $\epsilon > 0$ and then taking the limit $\epsilon \to 0$, you get that your desired limit is $0$.

You can also use Lebesgue's criteria for the Riemann integral instead of a proof with $\epsilon$.

Answer 2

It's not true.

Assuming $a > 1$, we can bound $$\frac1{1+x^n} \le \frac12, \forall n \in \mathbb{N}$$ which is an integrable function on $[1, a]$. On the other hand, since $x \ge 1$, we have that $$\frac1{1+x^n} \xrightarrow{n\to\infty}\begin{cases} 1, &\text{ if }x = 1 \\ 0, &\text{ if }x \in (1,a]\end{cases} $$

Hence $\left(\frac1{1+x^n}\right)_n$ converges almost everywhere to $0$, so Lebesgue dominated convergence theorem implies:

$$\lim_{n\to\infty} \int_1^a \frac{dx}{1+x^n} = \int_1^a \underbrace{\left(\lim_{n\to\infty} \frac1{1+x^n}\right)}_{=0} \,dx = 0$$

Answer 3

I will be computing $$J_n(a)=\int_1^a\ln(1+x^n)dx$$ First recall the definition of the complex root of a complex number $z$: $$z^{1/n}=|z|^{1/n}\exp\left[\frac{i}n(2\pi k+\arg z)\right],\qquad k=0,1,..,n-1$$ So if $x$ satisfies $1+x^n=0$ then $$x=\exp\frac{i\pi(2k+1)}n$$ So we define $$\lambda_{n,k}=\exp\frac{i\pi(2k+1)}n$$ And with this we can factor $x^n+1$: $$x^n+1=(x-\lambda_{n,0})(x-\lambda_{n,1})\cdots (x-\lambda_{n,n-1})=\prod_{k=0}^{n-1}(x-\lambda_{n,k})$$ And since $$\ln(xy)=\ln x+\ln y$$ We have that $$\ln(x^n+1)=\sum_{k=0}^{n-1}\ln(x-\lambda_{n,k})$$ So $$J_n(a)=\int_1^a\sum_{k=0}^{n-1}\ln(x-\lambda_{n,k})dx=\sum_{k=0}^{n-1}\int_1^a\ln(x-\lambda_{n,k})dx$$ $$J_n(a)=\sum_{k=0}^{n-1}\left[\ln\left|\frac{j(a-\lambda_{n,k})}{j(1-\lambda_{n,k})}\right|+\lambda_{n,k}-1\right]\qquad\text{where}\ j(z)=z^z$$ I hope this helps.