For
\(|m| \in\mathbf{N}\setminus\{ 1\} \), let
$$\begin{gathered} k^{(1)}(m,y): = \frac{\ln\ln A_{\xi,\alpha} (m) - \ln \ln [(y - \eta)(\cos\beta- 1)]}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} [(y - \eta)(\cos\beta- 1)]},\quad y < - \frac{3}{2}, \\ k^{(2)}(m,y): = \frac{\ln\ln A_{\xi,\alpha} (m) - \ln\ln[(y - \eta )(\cos\beta+ 1)]}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} [(y - \eta)(\cos\beta+ 1)]},\quad y > \frac{3}{2}. \end{gathered} $$
Then we have
$$k^{(1)}(m, - y) = \frac{\ln\ln A_{\xi,\alpha} (m) - \ln\ln[(y + \eta)(1 - \cos\beta)]}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} [(y + \eta)(1 - \cos\beta)]},\quad y > \frac{3}{2}, $$
yields
$$ \begin{aligned}[b] \omega(\lambda_{2},m) = {}&\sum _{n = - 2}^{ - \infty} \frac{k^{(1)}(m,n)\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{(n - \eta)(\cos \beta- 1)\ln^{1 - \lambda_{2}}[(n - \eta)(\cos\beta- 1)]} \\ &+ \sum_{n = 2}^{\infty} \frac{k^{(2)}(m,n)\ln^{\lambda_{1}}A_{\xi ,\alpha} (m)}{(n - \eta)(1 + \cos\beta)\ln^{1 - \lambda_{2}}[(n - \eta)(1 + \cos \beta)]} \\ ={}& \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 - \cos\beta} \sum_{n = 2}^{\infty} \frac{k^{(1)}(m, - n)}{(n + \eta)\ln^{1 - \lambda_{2}}[(n + \eta)(1 - \cos\beta)]} \\ &+ \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 + \cos\beta} \sum_{n = 2}^{\infty} \frac{k^{(2)}(m,n)}{(n - \eta)\ln^{1 - \lambda_{2}}[(n - \eta)(1 + \cos\beta)]}. \end{aligned} $$
(12)
In virtue of
\(0 < \lambda\le1\),
\(0 < \lambda_{2} < 1 \), and Example
1, we find that for
\(y > \frac{3}{2} \),
$$\begin{gathered} \frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta)\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta)(1 + ( - 1)^{i}\cos\beta)]} > 0, \\ \frac{d}{dy}\frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta)\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta)(1 + ( - 1)^{i}\cos\beta)]} < 0, \\ \frac{d^{2}}{dy^{2}}\frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta )\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta)(1 + ( - 1)^{i}\cos\beta )]} > 0\quad(i = 1,2), \end{gathered} $$
it follows that
$$\frac{k^{(i)}(m,( - 1)^{i}y)}{(y - ( - 1)^{i}\eta)\ln^{1 - \lambda_{2}}[(y - ( - 1)^{i}\eta)(1 + ( - 1)^{i}\cos\beta)]}\quad(i = 1,2) $$
are strictly decreasing and convex in
\(( \frac{3}{2},\infty )\). Then, by (
5), (
12) yields
$$\begin{aligned} \omega(\lambda_{2},m) < {}& \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 - \cos\beta} \int_{\frac{3}{2}}^{\infty} \frac{k^{(1)}(m, - y)}{(y + \eta)\ln^{1 - \lambda_{2}}[(y + \eta)(1 - \cos\beta)]} \,dy \\ &+ \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 + \cos\beta} \int_{\frac{3}{2}}^{\infty} \frac{k^{(2)}(m,y)}{(y - \eta)\ln^{1 - \lambda_{2}}[(y - \eta)(1 + \cos\beta)]} \,dy. \end{aligned} $$
Setting
\(u = \frac{\ln[(y + \eta)(1 - \cos\beta)]}{\ln A_{\xi,\alpha} (m)}\) (
\(u = \frac{\ln[(y - \eta)(1 + \cos\beta)]}{\ln A_{\xi,\alpha} (m)}\)) in the above first (second) integral, in view of Remark
1, we obtain
$$\begin{aligned} \omega(\lambda_{2},m) &< \biggl( \frac{1}{1 - \cos\beta} + \frac{1}{1 + \cos\beta} \biggr) \int_{0}^{\infty} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du \\ &= \frac{2\csc^{2}\beta}{\lambda^{2}} \int_{0}^{\infty} \frac{\ln v}{v - 1} v^{(\lambda_{2}/\lambda) - 1}\,dv = \frac{2\pi^{2}\csc^{2}\beta}{ \lambda^{2}\sin^{2}(\frac{\pi\lambda_{1}}{\lambda} )} = k_{\beta} (\lambda_{1}) \end{aligned} $$
by simplifications. Similarly, by (
5), (
12) also yields
$$\begin{aligned} \omega(\lambda_{2},m) >{}& \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 - \cos\beta} \int_{2}^{\infty} \frac{k^{(1)}(m, - y)}{(y + \eta)\ln^{1 - \lambda_{2}}[(y + \eta)(1 - \cos\beta)]} \,dy \\ &+ \frac{\ln^{\lambda_{1}}A_{\xi,\alpha} (m)}{1 + \cos\beta} \int_{2}^{\infty} \frac{k^{(2)}(m,y)}{(y - \eta)\ln^{1 - \lambda _{2}}[(y - \eta)(1 + \cos\beta)]} \,dy \\ \ge{}& \biggl(\frac{1}{1 - \cos\beta} + \frac{1}{1 + \cos\beta} \biggr) \int_{\frac{\ln[(2 + \eta)(1 + \cos\beta)]}{\ln A_{\xi,\alpha} (m)}}^{\infty} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du \\ ={}& k_{\beta} (\lambda_{1}) - 2\csc^{2}\beta \int_{0}^{\frac{\ln[(2 + \eta )(1 + \cos\beta)]}{\ln A_{\xi,\alpha} (m)}} \frac{\ln u}{u^{\lambda} - 1}u^{\lambda_{2} - 1} \,du \\ ={}& k_{\beta} (\lambda_{1}) \bigl(1 - \theta( \lambda_{2},m)\bigr) > 0, \end{aligned} $$
where
\(\theta(\lambda_{2},m)\) (<1) is indicated by (
11). Since
$$\frac{\ln u}{u^{\lambda} - 1}u^{\lambda_{2}/2} \to0\quad\bigl(u \to0^{ +} \bigr);\qquad \frac{\ln u}{u^{\lambda} - 1}u^{\lambda_{2}/2} \to\frac{1}{\lambda} \quad(u \to1), $$
there exists a positive constant
C such that
\(\frac{\ln u}{u^{\lambda} - 1}u^{\lambda_{2}/2} \le C\) (
\(0 < u \le1\)), and then for
\(A_{\xi,\alpha} (m) \ge(2 + \eta)(1 + \cos\beta)\), we have
$$ \begin{aligned}[b] 0 &< \theta(\lambda_{2},m) \le C\biggl[ \frac{\lambda}{\pi} \sin\biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln[(2 + \eta)(1 + \cos \beta )]}{\ln A_{\xi,\alpha} (m)}} u^{\frac{\lambda_{2}}{2} - 1} \,du \\ &= \frac{2C}{\lambda_{2}}\biggl[\frac{\lambda}{\pi} \sin\biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2}\biggl\{ \frac{\ln[(2 + \eta)(1 + \cos\beta )]}{\ln A_{\xi,\alpha} (m)}\biggr\} ^{\frac{\lambda_{2}}{2}}. \end{aligned} $$
(13)
Hence, (
10) and (
11) are valid. □