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\begin{equation}  s_0=-\int_{-\infty}^\infty \frac{1}{\sqrt{\pi}}e^{-x^2}\mathrm{ln}\left(\frac{1}{\sqrt{\pi}}e^{-x^2}\right)\;dx = \frac{1}{2}(\ln(\pi)+1)  \end{equation}  With added constants we find \begin{equation}  s_0=-\int_{-\infty}^\infty \frac{a}{\sqrt{\pi}}e^{-bx^2}\mathrm{ln}\left(\frac{a}{\sqrt{\pi}}e^{-bx^2}\right)\;dx = \frac{a}{2\sqrt{b}}(\ln(\frac{\pi}{a^2})+1)  \end{equation}  Then for the next integrals with $H_1(x)=2x$ we have \begin{equation}  s_1 = \frac{3\sqrt{\pi}}{4} -\int_{-\infty}^\infty e^{-x^2}x^2\ln(x^2) \; dx - \int_{-\infty}^\infty e^{-x^2}x^2\ln(A) \; dx