calculus - the value of $limlimits_{nrightarrowinfty}n^2left(int_0^1left(1+x^nright)^frac{1}{n} , dx-1right)$

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

Answer

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

$$\begin{align*} \lim_{n\to\infty} a_{n} = \int_{0}^{1} \int_{y}^{1} \frac{dtdy}{t(1+y)} &= - \int_{0}^{1} \frac{\log y}{1+y} \, dy \\ &= \sum_{m=1}^{\infty} (-1)^{m} \int_{0}^{1} y^{m-1} \log y \, dy = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^2} = \frac{\pi^2}{12}. \end{align*}$$