Unexpected result?

For all $n>1$ , if $f_n(x)=\sin\left(g_n(x)\right)$ such that $g_n(x)=x^n$ and $\alpha =\lim_{n\to \infty} n\left(\int_0^{\infty}\frac{f_n(x)}{g_n(x)}dx-1\right) ,$ then the following holds $\lim_{n\to \infty}n^2\left(-\frac{\alpha}{n}+\int_0^{\infty}\frac{f_n(x)}{g_n(x)}dx -1\right)=1-\gamma +\frac{\gamma^2}{2}-\frac{\pi^2}{a}$ for some positive integer $a$ . Find the value of $a^2$ .

Notation : $\gamma$ denotes Euler-Mascheroni constant.

Inspired by Aman Rajput's post.