## 1 Introduction
### Quasi-Classical Dielectric Relaxation Functions
For the diffusion relaxation domain, when inequality \(T_{\mathrm{n,D}}=\frac{T_{\mathrm{D}}}{\mathrm{n}^{2}}<T_{\mathrm{m}}\)_i_ \(\frac{T_{\mathrm{D}}}{T_{\mathrm{m}}}<\mathrm{n}^{2}\) holds, applying the following expressions is convenient:  
\[\Gamma^{(\omega)}(\mathrm{T})=\Gamma_{1}^{(\omega)}-\mathrm{i}\Gamma_{2}^{( \omega)} \tag{118}\]  
\[\Gamma_{1}^{(\omega)}(\mathrm{T})=\frac{8}{\pi^{2}}{\sum_{n=1}} \left[\frac{\frac{T_{\mathrm{n}}}{T_{\mathrm{m}}}\sin^{2}\left(\frac{\varpi }{2}\right)}{\mathrm{n}^{2}(1+\omega^{2}T_{\mathrm{n}}^{2})}\right]=\Gamma_{1,\mathrm{D}}^{(\omega)}(\mathrm{T})=\!\frac{4T_{\mathrm{D}}}{\pi^{2}T_{ \mathrm{M}}}{\sum_{n=1}}\!\left[\frac{\left(1-(-1)^{n}\right)\cdot\left( \mathrm{n}^{2}+\frac{T_{\mathrm{D}}}{T_{\mathrm{M}}}\right)}{\mathrm{n}^{2} \!\left(\left(\mathrm{n}^{2}+\frac{T_{\mathrm{D}}}{T_{\mathrm{m}}}\right)^{2 }+\omega^{2}T_{\mathrm{D}}^{2}\right)}\right], \tag{118a}\]  
\[\Gamma_{2}^{(\omega)}(\mathrm{T})=\frac{8}{\pi^{2}}{\sum_{n=1}} \!\left[\frac{\frac{\omega\tau_{n}^{2}}{T_{\mathrm{M}}}\sin^{2}\left(\frac{ \varpi}{2}\right)}{\mathrm{n}^{2}(1+\omega^{2}T_{\mathrm{n}}^{2})}\right]\! =\Gamma_{2\mathrm{D}}^{(\omega)}=\frac{4T_{\mathrm{D}}}{\pi^{2}T_{ \mathrm{M}}}{\sum_{n=1}}\!\left[\frac{\left(1-(-1)^{n}\right)\cdot\omega T_{ \mathrm{D}}}{\mathrm{n}^{2}\!\left(\left(\mathrm{n}^{2}+\frac{T_{\mathrm{D}}} {T_{\mathrm{M}}}\right)^{2}+\omega^{2}T_{\mathrm{D}}^{2}\right)}\right], \tag{118b}\]  
which corresponds to generalized equality:  
\[\Gamma_{1}^{(\omega)}(\mathrm{T})=\frac{4\alpha_{1}}{\pi^{2}}{\sum_{n=1}} \left[\frac{\left(1-(-1)^{n}\right)\cdot\left(\mathrm{n}^{2}+\alpha_{1} \right)}{\mathrm{n}^{2}\!\left(\left(\mathrm{n}^{2}+\alpha_{1}\right)^{2}+ \alpha_{2}^{2}\right)}\right], \tag{119a}\]  
\[\Gamma_{2}^{(\omega)}\left(\mathrm{T}\right)=\frac{4\alpha_{1}\alpha_{2}}{\pi ^{2}}\sum_{n=1}^{\infty}\left[\frac{\left(1-(-1)^{n}\right)}{\mathrm{n}^{2} \!\left(\left(\mathrm{n}^{2}+\alpha_{1}\right)^{2}+\alpha_{2}^{2}\right)}\right] \tag{119b}\]  
and, accordingly, the following identity:  
\[ \begin{array}{l}\Gamma_{1}^{(\omega)}(\mathrm{T})=\Gamma_{1,\mathrm{D}}^{( \omega)}(\mathrm{T})=\frac{4\alpha_{1}}{\pi^{2}}\!\left(\Gamma_{11}^{(\omega) }(\mathrm{T})+\alpha_{1}\Gamma_{12}^{(\omega)}(\mathrm{T})\right)\!,\Gamma_ {2}^{(\omega)}(\mathrm{T})=\Gamma_{2,\mathrm{D}}^{(\omega)}(\mathrm{T})=\\ \frac{4\alpha_{1}\alpha_{2}}{\pi^{2}}\times\times\Gamma_{12}^{(\omega)}( \mathrm{T}).\end{array}  \tag{120}\]  
Computing the series provides us with the following:  
\[ \begin{array}{l}\Gamma_{11}^{(\omega)}(\alpha_{1},\alpha_{2})=\sum\limits_{ n=1}^{\infty}\left[\frac{1-(-1)^{n}}{\left(\mathrm{n}^{2}+\alpha_{1}\right)^{2}+ \alpha_{2}^{2}}\right]=\\ =\frac{1}{2\alpha_{2}\sqrt{\alpha_{1}^{2}+\alpha_{2}^{2}}}\times\frac{\mathrm{ A}_{1}\mathrm{sh}(2\alpha_{2})-\Delta_{2}\cdot\mathrm{sin}(2\Delta_{1})}{ \mathrm{ch}^{2}(\Delta_{2})\cos^{2}(\Delta_{1})+\mathrm{sh}^{2}(\Delta_{2}) \sin^{2}(\Delta_{1})},\end{array}  \tag{121}\]  
\[ \begin{array}{l}\Gamma_{12}^{(\omega)}(\alpha_{1},\alpha_{2})=\sum\limits_ {n=1}^{\infty}\left[\frac{1-(-1)^{n}}{\mathrm{n}^{2}\left(\left(\mathrm{n}^{2} +\alpha_{1}\right)^{2}+\alpha_{2}^{2}\right)}\right]=\\ =\frac{\pi^{2}}{4\left(\alpha_{1}^{2}+\alpha_{2}^{2}\right)}\times\left[1- \frac{2(\alpha_{1}(\Delta_{1}\mathrm{sh}(2\alpha_{2})-\Delta_{2}\mathrm{sin}(2 \Delta_{1}))+\alpha_{2}(\Delta_{1}\mathrm{sin}(2\Delta_{1})+\Delta_{2} \mathrm{sh}(\Delta_{2})))}{\alpha_{2}\pi^{2}(\mathrm{ch}^{2}(\Delta_{2})\cos^ {2}(\Delta_{1})+\mathrm{sh}^{2}(\Delta_{2})\sin^{2}(\Delta_{1}))\sqrt{\alpha_{ 1}^{2}+\alpha_{2}^{2}}}\right].\end{array}  \tag{122}\]  
Here, \(\Delta_{1}=\frac{\pi}{2}\sqrt{\frac{\sqrt{\alpha_{1}^{2}+\alpha_{2}^{2}-\alpha_ {1}}}{2}}\) and \(\Delta_{2}=\frac{\pi}{2}\sqrt{\frac{\sqrt{\alpha_{1}^{2}+\alpha_{2}^{2}+\alpha_ {1}}}{2}}\).