How to calculate $\lim_{n\rightarrow\infty} n\int^{1}_{0}\ln(1+x^n)dx$? How to calculate the following limit? $$\lim_{n\rightarrow\infty} n\int^{1}_{0}\ln(1+x^n)dx.$$
It is possible to use the criterion sandwich?
Using inequality$$\ln(1+t)\leq t, t>-1$$ follows easily that$$\int^{1}_{0}\ln(1+x^n)dx <\int^{1}_{0}x^ndx =\frac{1}{n+1}$$ and consequently$$n\int^{1}_{0}\ln(1+x^n)dx<\frac{n}{n+1}.$$
Might minor integral in the same way?
 A: Let $t=x^n$ hence $x=t^{1/n}$ and $dx=\frac 1 n t^{1/n-1}dt$ so
$$n\int_0^1\ln(1+x^n)dx=\int_0^1\ln(1+t)t^{1/n-1}dt$$
so by the dominated convergence theorem we have
$$\lim_{n\rightarrow\infty} n\int^{1}_{0}\ln(1+x^n)dx=\int_0^1\frac{\ln(1+t)}{t}dt=\int_0^1\sum_{n=1}^\infty(-1)^{n-1}\frac{t^{n-1}}{n}dt=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2} $$
and knowing that
$$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$
we find easily that $\frac{\pi^2}{12}$ is the desired result.
A: Caveat:  I'm going to play fast and loose with convergence issues, so the following won't be a rigorous proof.
Expand $\ln(1+x^n)$ as its Taylor series, obtaining $$\int_0^1\ln(1+x^n)\ dx = \int_0^1 \sum_{k=1}^\infty (-1)^{k-1}\frac{x^{nk}}{k}\ dx = \sum_{k=1}^\infty (-1)^{k-1}\frac{1}{k(kn+1)}.$$
As $n\rightarrow\infty$, $n$ times this sum becomes $$n\sum_{k=1}^\infty (-1)^{k-1}\frac{1}{k(kn+1)} \longrightarrow \sum_{k=1}^\infty (-1)^{k-1}\frac{1}{k^2} = \sum_{k=1}^\infty \frac{1}{k^2} - 2\sum_{k=1}^\infty\frac{1}{4k^2} = \frac{\pi^2}{6}- 2\frac{\pi^2}{24} = \frac{\pi^2}{12}.$$
