I am trying the series expansion of this function : $$f(n)=\, _2F_1\left(-\frac{1}{n},\frac{1}{n};\frac{1}{n}+1;-1\right)\text{at}\; \infty$$
This comes from : find an asymptotic series for this integral
$$I(n)=\int_{0}^{1}\sqrt[n]{1+x^n}dx$$ By using Mathematica, I could find $$I(n)=1+\frac{\pi^2}{12n^2}-\frac{5\zeta(3)}{8n^3}+\frac{x}{n^4}+O\left(\frac{1}{n^5}\right)$$
But I can't find what is $x$ equals to. So, I want to ask how to get that asymptotic series and can we use elementary method to find an asymptotic series for this integral? Thank you for your time and your help.
By performing the change of integration variables from $x$ to $t$ via $x^n = t$, followed by integration by parts, we arrive at $$ \int_0^1 {(1 + x^n )^{1/n} \mathrm{d}x} = \frac{1}{n}\int_0^1 {(1 + t)^{1/n} t^{1/n - 1} \mathrm{d}t} = 2^{1/n} - \frac{1}{n}\int_0^1 {(1 + t)^{1/n - 1} t^{1/n} \mathrm{d}t}. $$ Now, observe that $$ 2^{1/n} = \sum\limits_{k = 0}^\infty {\frac{{\log ^k 2}}{{k!}}\frac{1}{{n^k }}} $$ and $$ (1 + t)^{1/n - 1} t^{1/n} = \sum\limits_{k = 0}^\infty {\frac{1}{{k!}}\frac{{\log ^k ((1 + t)t)}}{{1 + t}}\frac{1}{{n^k }}} $$ for any $n \in \mathbb{Z}^+$ and $0 < t < 1$. Consequently, we can expand $I(n)$ as $$ I(n) = 1 + \sum\limits_{k = 1}^\infty \frac{a_k}{n^k} $$ for any $n \in \mathbb{Z}^+$, with $$ a_k=\frac{\log ^k 2}{k!} - \frac{1}{(k - 1)!}\int_0^1 \frac{\log ^{k - 1} ((1 + t)t)}{1 + t} \mathrm{d}t. $$ The coefficients $a_k$ can be represented using polylogarithms and values of the Riemann zeta function, thus yielding the coefficients as provided in the comments.
To list the first five coefficients, you can use the following Wolfram Mathematica commands: