e^{2\nu^2} -
\frac{1}{\nu^4} e^{\nu^2}\,$$ and the matching condition $v_{\nu, 1}'(\frac{1}{\nu}) =
v_{\nu, 2}'(\frac{1}{\nu})$ reads $$\frac{1}{\nu} =
\nu^2 \alpha_\nu e^{\nu} + \nu^2 \beta_\nu e^{-\nu} \, ,$$ leading to $$\alpha_\nu = \frac{1}{\nu^4 e^{\nu^2}}
\frac{1 + \nu e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}}\,.$$ Substituting this value for $\alpha_\nu$ into the formula for $\beta_{\nu}$ one obtains $$\beta_\nu =
\frac{e^\nu}{\nu^3}
\frac{1 - \frac{1}{\nu} e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}}\,.$$ It remains to determine $\gamma_\nu$ from the matching condition $v_{\nu, 1}(\frac{1}{\nu}) = v_{\nu, 2}(\frac{1}{\nu})$. One has $$- \frac{1}{2 \nu^2} + \gamma_\nu =
\frac{1}{\nu^4}  - \alpha_\nu e^{\nu}
+ \beta_\nu e^{-\nu}\, ,$$ which gives the following formula for $\gamma_\nu$ $$\gamma_\nu = \frac{1}{2 \nu^2} + \frac{1}{\nu^4}
+ \frac{1}{\nu^3} \frac{1 - \frac{1}{\nu} e^{-\nu^2 +\nu}}
{1 + e^{-2\nu^2 + 2\nu}}
- \frac{1}{\nu^4 e^{\nu^2 - \nu}}
\frac{1 + \nu e^{-\nu^2 +\nu}}{1 + e^{-2\nu^2 + 2\nu}} \, .$$ Now let us turn to the estimate of the quotient, $\frac{\int_{1/\nu}^1 v_\nu}{\int_{0}^{1/\nu} v_\nu}$. We estimate the two integrals separately. First we derive a lower bound for $\int_{0}^{1/\nu} v_\nu$. To this end note that $$\int_{0}^{1/\nu} v_\nu =
\bigg( -\frac{1}{6}x^3 + \gamma_\nu x \bigg)\bigg|_0^{1/\nu}
= \frac{\gamma_\nu}{\nu} - \frac{1}{6\nu^3}$$ yielding $$\int_{0}^{1/\nu} v_\nu =
\frac{1}{3 \nu^3}
+ \frac{1}{\nu^4} \frac{1 - \frac{1}{\nu} e^{-\nu^2 +\nu}}
{1 + e^{-2\nu^2 + 2\nu}}
+ 