It's an exercise, I need to find an equivalent of :
$$\int_{1}^{+\infty}\exp{(-x^n)}dx$$
I tried this : Let $x=u^{1/n}.$ Then $dx=\frac{1}{n}u^{1/n-1}\,du,$ so : \begin{align} \int_1^{+\infty} e^{-x^n}\,dx & =\int_0^{+\infty} e^{-x^n}\,dx-\int_0^1 e^{-x^n}\,dx=\frac{1}{n}\int_0^{+\infty} u^{1/n-1}e^{-u}\,du-\int_0^1 e^{-x^n}\,dx\\ & = \frac{1}{n}\Gamma(1/n)-\int_0^1 e^{-x^n}\,dx = \frac{1}{n}(n+o(n))-1+o(1)\\ & = o(1) \end{align}
I try to multiply by $n$ but I am stuck here. One person told me that there will be a euler constant after that (I doubt it). Yet I don't know how to show it.
I would like first prove that there is a constant $c\neq 0$ such that this integral multiply by $n$ equals to $c$. Then try to find it.
Using the incomplete gamma function and the exponential integral function $$I_n=\int\exp{(-x^n)}\,dx=-\frac{x }{n}E_{\frac{n-1}{n}}\left(x^n\right)$$
$$J_n=\int_{1}^{\infty}\exp{(-x^n)}\,dx=\frac{1}{n}\Gamma \left(\frac{1}{n},1\right)$$
If we expand $\Gamma \left(\frac{1}{n},1\right)$ as a series for large value of $n$, the constant term is $c_0=-\text{Ei}(-1)$ as already given by @Sangchul Lee in comments.
What is difficult is to simplify the next since they involve more and more complex Meijer G-functions such as $$c_1=G_{2,3}^{3,0}\left(1\left| \begin{array}{c} 1,1 \\ 0,0,0 \end{array} \right.\right)$$ $$c_2=G_{3,4}^{4,0}\left(1\left| \begin{array}{c} 1,1,1 \\ 0,0,0,0 \end{array} \right.\right)$$ and so on. Less compact forms involve hypergeometric functions.
Making their decimal values rational $$n\,J_n=-\text{Ei}(-1)+\frac{59}{603\,n}+\frac{23}{646\,n^2}+\frac{3}{271\,n^3}+O\left(\frac{1}{n^4}\right)$$
For $n=10$, the above would give $J_{10}=0.0229524$ while the exact value is $0.0229536$