\frac{1}{\nu^5}\Big( 1 -  \frac{1}{ e^{\nu^2 - \nu}}
\frac{1 + \nu e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}} \Big).$$ Note that for any $\nu > 1$, one has $1 - \frac{1}{\nu} e^{-\nu^2 +\nu} > 0,$ and $$\frac{1}{ e^{\nu^2 - \nu}}
\frac{1 + \nu e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}}
\le 1\, ,$$ implying that $$\label{estimate integral in denominator}
\int_{0}^{1/\nu} v_\nu  \ge \frac{1}{3\nu^3} \, , \quad
\forall \nu > 1\, .$$ Next we derive an upper bound for $\int_{1/\nu}^1 v_\nu$. One computes $$\begin{aligned}
\label{t1}
\int_{1/\nu}^1 v_\nu &=
\frac{1}{\nu^4} (1 - \frac{1}{\nu})
- \frac{\alpha_\nu}{\nu^2} e^{\nu^2 x}|_{1/\nu}^1
- \frac{\beta_\nu}{\nu^2} e^{-\nu^2 x}|_{1/\nu}^1 \nonumber \\
& \le
\frac{1}{\nu^4}
+ \frac{\alpha_\nu}{\nu^2} e^{\nu}
+ \frac{\beta_\nu}{\nu^2} e^{-\nu} \nonumber \\
& \le \frac{1}{\nu^4}  ( 1+ R_\nu ),\,\end{aligned}$$ where the remainder $R_\nu$ is given, and bounded, by $$\begin{aligned}
\label{t2}
R_\nu &= \frac{1}{\nu^2 e^{\nu^2 - \nu}}
\frac{1 + \nu e^{-\nu^2}}{1 + e^{-2\nu^2 + 2\nu}}
+ \frac{1}{\nu}
\frac{1 - \frac{1}{\nu} e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}}\nonumber \\ &
 =\frac{1+e^{-2\nu^2 + \nu}}{\nu\big(1+e^{-2\nu^2 + 2\nu}\big)}\nonumber \\ &\le \frac{1}{\nu},\qquad \forall\nu>1.\end{aligned}$$ By and , $$\label{estimate integral in nominator}
\int_{1/\nu}^1 v_\nu \le \frac{1}{\nu^4}+\frac{1}{\nu^5},\qquad \forall\nu\ge 1.$$ Combining and we obtain the claimed estimate , $$\frac{\int_{1/\nu}^1 v_\nu}{\int_{0}^{1/\nu} v_\nu}
\le \frac{3}{\nu}+\frac{3}{\nu^2}\le \frac{6}{\nu},\, \forall\nu\ge 1.$$ The remaining statements of the theorem now follow 