if $n=1$, \begin{eqnarray*} \int^{+\infty}_1 \frac{\widehat{A_1}^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau&=&\bigg[-n\tau^{-\frac{1}{n}}\widehat{A_1}^{-1}(\tau)\bigg]^{+\infty}_1\\ &+&n\int^{+\infty}_1\frac{\widehat{A}^{-1}(\tau)}{\tau^{1+\frac{2}{n}}}d\tau\\ &\leq& \widehat{A_1}(1)+\int^{+\infty}_1\frac{A^{-1}}{\tau^3}d\tau\\ &\leq& \widehat{A_1}^{-1}(1)+\int^{+\infty}_1\frac{1}{\tau^2}d\tau\\ \end{eqnarray*} $\Rightarrow q(A,1)=1$
if $n>1$ we suppose $q>n-1$ we notice that \begin{eqnarray*} \widehat{A_2}^{-1}(t)&=&\int^t_0 \frac{\widehat{A_1}^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau\\ &=& \int^1_0 \frac{\widehat{A_1}^{-1}(\tau)}{\tau^{1+\frac{1}{n}}}d\tau+\bigg[-n\tau^{-\frac{1}{n}} \widehat{A_1}^{-1}(\tau) \bigg]^t_1\\ &+&n\int^t_1\frac{\widehat{A}^{-1}(\tau)}{\tau^{1+\frac{2}{n}}}d\tau\\ \end{eqnarray*}
$$ \widehat{A_2}(t)\leq k_2\int^t_0\frac{\widehat{A}^{-1}}{\tau^{1+\frac{2}{n}}}d\tau $$
similarly ,we find $$ \widehat{A_3}^{-1}(t)\leq K_3 \int^t_0\frac{\widehat{A}^{-1}(\tau)}{\tau^{1+\frac{3}{n}}}d\tau $$ until n we find $$ \widehat{A_n}^{-1}(t)\leq k_n \int^t_0\frac{\widehat{A}^{-1}}{\tau^2}d\tau$$ then \begin{eqnarray*} \int^{+\infty}_1 \frac{\widehat{A_n}^{-1}(t)}{t^{1+\frac{1}{n}}}dt &\leq& k_n \bigg[-nt^{-\frac{1}{n}}\int^t_0\frac{\widehat{A}^{-1}(\tau)}{\tau^2}d\tau \bigg]^{+\infty}_1\\ &+&n\int^{+\infty}_1\frac{\widehat{A}^{-1}}{\tau^{2+\frac{1}{n}}}d\tau\\ &\leq& k_{n+1}+n\int^{+\infty}_1\frac{1}{t^{1+\frac{1}{n}}}d\tau\\ &\leq& k_{n+1}+n^2\\ \end{eqnarray*}
so $$q(a,n)\leq n$$$$q(A,n)\leq n$$