# Prove that $\lim n\int_1^a\frac{1}{1+x^n}dx=\ln 2$

My problem is that for a given $$a>1$$, we have that $$\lim_{n\to\infty}n\int_{1}^{a}\frac{1}{1+x^n}dx=\ln 2$$

The natural idea seems to be to add and substract $$x^n$$ from the numerator and we obtain easily that $$n\int_{1}^{a}\frac{1}{1+x^n}dx=n(a-1)-a\ln(1+a^n)+\ln2+\int_1^a\ln(1+x^n)dx$$ which would sort of explain the $$\ln 2$$ result but I can't continue from here.

Let $$I_n =n\int_1^a\frac{1}{1+x^n}\mathrm{d}x.$$By making change of variables $$x = 1+\frac{u}{n}$$, we obtain $$I_n =\int_0^{n(a-1)}\frac{1}{1+\left(1+\frac{u}{n}\right)^n}\mathrm{d}u=\int_0^{\infty}\frac{1_{\{u\le n(a-1)\}}}{1+\left(1+\frac{u}{n}\right)^n}\mathrm{d}u.$$ Since $$\left(1+\frac{u}{n}\right)^n\le \left(1+\frac{u}{n+1}\right)^{n+1} \longrightarrow e^u$$ for all $$u\ge 0$$, we find $$\frac{1_{\{u\le n(a-1)\}}}{1+\left(1+\frac{u}{n}\right)^n}\le \frac{1}{1+\left(1+\frac{u}{2}\right)^2},\quad\forall n\ge 2.$$ The RHS is an integrable function, hence by dominated convergence theorem it follows $$\begin{eqnarray} \lim_{n\to\infty} I_n& =&\int_0^\infty \frac{1}{1+e^u}\mathrm{d}u\\&=&\int_1^\infty \frac{1}{s(s+1)}\mathrm{d}s\tag{e^u=s}\\&=&\lim_{n\to\infty}\int_1^n \left(\frac{1}{s}-\frac{1}{s+1}\right)\mathrm{d}s\\ &=&\int_1^2 \frac{1}{s}\mathrm{d}s-\lim_{n\to\infty}\int_{n}^{n+1} \frac{1}{s}\mathrm{d}s\\ &=&\int_1^2 \frac{1}{s}\mathrm{d}s=\ln 2. \end{eqnarray}$$
I didn't see Song's post until after I posted, so I deleted mine. Then I realized that this answer was a bit shorter, and possibly a bit easier to follow. \begin{align} \lim_{n\to\infty}n\int_1^a\frac1{1+x^n}\,\mathrm{d}x &=\lim_{n\to\infty}\int_1^\infty[x\le a^n]\frac{x^{\frac1n-1}}{1+x}\,\mathrm{d}x\tag1\\ &=\int_1^\infty\frac1{x(1+x)}\,\mathrm{d}x\tag2\\ &=\left.\log\left(\frac{x}{x+1}\right)\right]_1^\infty\tag3\\[6pt] &=\log(2)\tag4 \end{align} Explanation:
$$(1)$$: substitute $$x\mapsto x^{1/n}$$
$$(2)$$: dominated convergence; dominated by $$\frac{x^{-1/2}}{1+x}$$
$$(3)$$: integrate
$$(4)$$: evaluate