the value of $\lim\limits_{n\rightarrow\infty}n^2\left(\int_0^1\left(1+x^n\right)^\frac{1}{n} \, dx-1\right)$ This is exercise from my lecturer, for IMC preparation. I haven't found any idea.
Find the value of
$$\lim_{n\rightarrow\infty}n^2\left(\int_0^1 \left(1+x^n\right)^\frac{1}{n} \, dx-1\right)$$
Thank you
 A: By integration by parts,
\begin{align*}
\int_{0}^{1} (1 + x^{n})^{\frac{1}{n}} \, dx
&= \left[ -(1-x)(1+x^{n})^{\frac{1}{n}} \right]_{0}^{1} + \int_{0}^{1} (1-x)(1 + x^{n})^{\frac{1}{n}-1}x^{n-1} \, dx \\
&= 1 + \int_{0}^{1} (1-x) (1 + x^{n})^{\frac{1}{n}-1} x^{n-1} \, dx
\end{align*}
so that we have
\begin{align*}
n^{2} \left( \int_{0}^{1} (1 + x^{n})^{\frac{1}{n}} \, dx - 1 \right)
&= \int_{0}^{1} n^{2} (1-x) (1 + x^{n})^{\frac{1}{n}-1} x^{n-1} \, dx.
\end{align*}
Let $a_{n}$ denote this quantity. By the substitution $y = x^{n}$, it follows that
\begin{align*}
a_{n}
&= \int_{0}^{1} n \left(1-y^{1/n}\right) (1 + y)^{\frac{1}{n}-1} \, dy
= \int_{0}^{1} \int_{y}^{1} t^{\frac{1}{n}-1} (1 + y)^{\frac{1}{n}-1} \, dtdy
\end{align*}
Since $0 \leq t (1 + y) \leq 2$ and $ \int_{0}^{1} \int_{y}^{1} t^{-1}(1+y)^{-1} \, dtdy < \infty$, an obvious application of the dominated convergence theorem shows that
\begin{align*}
\lim_{n\to\infty} a_{n}
 = \int_{0}^{1} \int_{y}^{1} \frac{dtdy}{t(1+y)}
&= - \int_{0}^{1} \frac{\log y}{1+y} \, dy \\
&= \sum_{m=1}^{\infty} (-1)^{m} \int_{0}^{1} y^{m-1} \log y \, dy
 = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^2}
 = \frac{\pi^2}{12}.
\end{align*}
A: Mathematica evaluates the integral to
$$\int_0^{1}(1+x^n)^{1/n}dx={}_2F_1\left(-\frac{1}{n},\frac1n,1+\frac1n;-1\right).\tag{1}$$
Next, let us write the standard series representation for the hypergeometric function
$$_2F_1(a,b,c;t)=\sum_{k=0}^{\infty}\alpha_kt^k,\qquad \alpha_k=\frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c)}{k!\,\Gamma(a)\Gamma(b)\Gamma(c+k)}.$$
Now an easily verified claim: as $n\rightarrow\infty$, for the parameters as in (1), we have $\alpha_0=1$ and
$$\alpha_k\sim-\frac{1}{k^2n^2}+O(n^{-3}).$$
Hence we obtain
\begin{align}
\lim_{n\rightarrow\infty}n^2\left(\int_0^{1}(1+x^n)^{1/n}dx-1\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}=\frac{\pi^2}{12},
\end{align}
which is also confirmed by numerical calculation with $n\sim 10^4-10^8$.
