TITLE: Prove that $\lim n\int_1^a\frac{1}{1+x^n}dx=\ln 2$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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