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When solving for the next integrals we have $H_2(x)=4x^2-2$, and additional complexity. We find by laking a log expansion some progress can be made to finding a closed form for the total entropy of the density state.  We have \begin{equation}  -\int_{-\infty}^{\infty}\frac{e^{-x^2}(16x^4-16x^2+4)}{8\sqrt{\pi}}\left(\sum_{n=1}^\infty s_2=-\int_{-\infty}^{\infty}\frac{e^{-x^2}(16x^4-16x^2+4)}{8\sqrt{\pi}}\left(\sum_{n=1}^\infty  \frac{1}{n}\left(\frac{e^{-x^2}(16x^4-16x^2+4)}{8\sqrt{\pi}} -1 \right)^n \right) \right)\;dx  \end{equation}