$\lambda_n$ and the inequality follow from and if we put $m = q + 1$ in and notice that $2 \sqrt{2} < \pi$ when we multiply the constant factors in inequalities.

Analysis of the function N sub \* alpha.
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This function is determined by the inequality . Later we consider potentials with the coupling coefficient $s$ \[see , \] so it is useful to know the behavior of $X = X_{\beta}(t)$, the solution of the equation $$t \frac{\log eX}{X^\beta} = \frac{1}{2}, \quad \beta = 2\alpha, \text{ for large }t. \label{eq:tfromndef}$$ Let us rewrite it as $$\begin{aligned}
\tau \log Y = Y, \, \, & \text{where } Y = (eX)^{\beta}  \label{eq:tfromndef2.1}\\
& \text{and } \tau = \frac{2}{\beta} e^{\beta}t \label{eq:tfromndef2.2}\end{aligned}$$ The equation has solutions $1 < y < Y$, such that $$\begin{aligned}
y(\tau) & = 1 + \frac{1}{\tau} + O {\left( \frac{1}{\tau^2} \right)}, &&\tau \to \infty \label{eq:y1def}\\
\intertext{and}
Y(\tau) & \to \infty, && \tau \to \infty. \label{eq:y2def}\end{aligned}$$

The solution $Y$ is $$\begin{aligned}
Y(\tau) &= \tau \log \tau \cdot (1 + r(\tau)) \label{eq:y2behave}\\
\intertext{where}
r(\tau) &= \frac{\log \log \tau}{\log \tau} \left( 1 + o(1) \right) \label{eq:y2r}\\
\intertext{so for any $\delta$ we can find $\tau^*$ such that}
Y(\tau) &\leq \tau \log \tau + 