Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$ 
Find the value of the limit $$\lim_{n \to \infty}n\left(\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$$

I can't solve the integral $\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x$. But maybe the questions doesn't require solving the integral. 
Apparently the $\lim_{n \to \infty}(\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x)^n$ should be $\frac{1}{2}$ for the question to make sense. That's all I know.
 A: Hint. If one knows the digamma function $\psi=\Gamma'/\Gamma$, one may write, as $n \to \infty$,
$$
\begin{align}
\int_0^1\frac1{1+x^n}\:dx&=\int_0^1\frac{1-x^n}{1-x^{2n}}\:dx
\\\\&\stackrel{u=x^{2n}}=\frac1{2n}\int_0^1\frac{(1-u^{1/2})u^{1/(2n)-1}}{1-u}\:du
\\\\&=\frac1{2n}\left[\psi\left(\frac1{2n}+\frac12 \right)-\psi\left(\frac1{2n} \right) \right]
\\\\&=1-\frac{\ln 2}{n}+\frac{\pi ^2-6 \ln^2 2}{12n^2}+O\left(\frac{1}{n^3}\right)
\end{align}
$$ giving, as $n \to \infty$,
$$
n\ln \left(1-\frac{\ln 2}{n}+\frac{\pi ^2-6 \ln^2 2}{12n^2}+O\left(\frac{1}{n^3}\right) \right)=-\ln 2+\frac{\pi ^2-6 \ln^2 2}{24n}+O\left(\frac{1}{n^2}\right)
$$ 
that is

$$
n\left[\left(\int_0^1\frac1{1+x^n}\:dx \right)^n-\frac12\right]\to \frac{\pi ^2-6 \ln^2 2}{24}.
$$

A: My approach is a little different.  I would sub $x=u^{1/n}$ in the integral and get
$$I(n) = \int_0^1 \frac{dx}{1+x^n} = \frac1n \int_0^1 du \frac{u^{\frac1n-1}}{1+u} = \frac1n \int_0^1 du \, u^{1/n} \left (\frac1u - \frac1{1+u} \right ) \\= 1-\frac1n \int_0^1 du \frac{u^{1/n}}{1+u}$$
We then note that $u^{1/n} = e^{\log{u}/n}$ and that $n$ is large enough for the following series expansion to be valid:
$$I(n) = 1- \sum_{j=0}^{\infty} \frac1{n^{j+1}} \int_0^1 du \frac{\log^j{u}}{1+u} $$
Due to the nature of the limit we are posed, we go out to second order; thus
$$I(n) = 1-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} +O \left ( \frac1{n^3} \right ) $$
Then
$$\begin{align}I(n)^n &= e^{n \log{\left [1-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} +O \left ( \frac1{n^3} \right )\right ]} } \\ &= e^{n\left [-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} - \frac{\log^2{2}}{2 n^2}+O \left ( \frac1{n^3} \right ) \right ] } \\ &= \frac12 \left [1+\left (\frac{\pi^2}{12} - \frac12 \log^2{2} \right ) \frac1n +O \left ( \frac1{n^2} \right ) \right ]\end{align}$$
The sought after limit is then

$$\lim_{n \to \infty} n \left [I(n)^n-\frac12 \right ] = \frac{\pi^2}{24} - \frac14 \log^2{2} $$

A: Let $I(n)$ be given by the integral 
$$\begin{align}
I(n)&=\int_0^1 \frac{1}{1+x^n}\,dx \tag 1\\\\
\end{align}$$
Then, expanding the integrand of the integral on the right-hand side of $(1)$ in the Taylor series $ \frac{1}{1+x^n}=\sum_{k=0}^\infty (-1)^kx^{nk}$ reveals
$$\begin{align}
I(n)&=\sum_{k=0}^\infty \frac{(-1)^k}{nk+1}\\\\
&=1+\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} \tag 2
\end{align}$$

Next, using the fact that $\log(2)= \sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$ and that $\frac{\pi^2}{12}=-\sum_{k=1}^\infty \frac{(-1)^k}{k^2}$, we can write the series in $(2)$ as
$$\begin{align}
\sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} &=-\log(2)+\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k+1/n}-\frac1k\right)\\\\
&=-\log(2)-\frac1n \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1/n)}\\\\
&=-\log(2)-\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k^2}-\frac1n\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k(k+1/n)}-\frac{1}{k^2}\right)\\\\
&=-\log(2)+\frac{\pi^2}{12n}+O\left(\frac1{n^2}\right) \tag 3
\end{align}$$

Using $(1)-(3)$ yields
$$I(n)=1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac1{n^3}\right) \tag 4$$

Next, using $(4)$, we can write
$$\begin{align}
\left(I(n)\right)^n&=e^{n\log\left(1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac1{n^3}\right)\right)}\\\\
&=e^{-\log(2)+\frac{\pi^2}{12n}-\frac{\log^2(2)}{2n}+O\left(\frac{1}{n^2}\right)}\\\\
&=\frac12 \left(1+\frac{\pi^2}{12n}-\frac{\log^2(2)}{2n}+O\left(\frac{1}{n^2}\right)\right)
\end{align}$$
Finally, we have

$$\begin{align}
\lim_{n\to \infty}\left(n\left(\left(I(n)\right)^n-\frac12\right)\right)&=\lim_{n\to \infty}\left(\frac{\pi^2}{24}-\frac{\log^2(2)}{4}+O\left(\frac1n\right)\right)\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\pi^2}{24}-\frac{\log^2(2)}{4}}
\end{align}$$

And we are done!
