18 scratch area
We have the identity \begin{gather} c(t)=\int_{E_0}^{\infty}dE \rho(E) e^{-Et} =\cancel{\rho\frac{e^{-Et}}{-t}\bigg|_{E_0}^{\infty}} -\int_{E_0}^{\infty}dE \rho'(E)\frac{e^{-Et}}{-t} \end{gather} thus using as a base \tilde b_t(E)=\frac{e^{-Et}}{t} to approximate \Delta using the HLT we can compute the convolution \int\rho'\Delta \begin{gather} \sum_t g_tc(t)=\int_{E_0}^{\infty}dE \rho'(E) \sum_t g_t\frac{e^{-Et}}{t}= \int_{E_0}^{\infty}dE \rho'(E) \tilde \Delta(E) \\=\cancel{\rho \Delta\bigg|_{E_0}^{\infty} }-\int_{E_0}^{\infty}dE \rho(E) \tilde \Delta'(E)\,. \end{gather} So if we use as base \tilde b_t(E)=\frac{e^{-Et}}{t} to approximate \Delta with the HLT we can compute the convolution -\int\rho\Delta'. For Z_0 the kernel function is \begin{gather} \theta_0(\omega-\omega_0)=\frac{c}{1+e^{\frac{\omega-\omega_0}{\sigma}}}\,,\\ -\int \theta_0(\omega-\omega_0)= c\left[\sigma \log\left(e^{\frac{\omega_0}{\sigma}}+e^{\frac{\omega}{\sigma}}\,. \right)-\omega\right] \end{gather}
One could also try -\int \theta_0(\omega-\omega_0)\approx-(\omega-\omega_0)\theta_0(\omega-\omega_0)
We could iterate the procedure and use as base b_t''=\frac{e^{-Et}}{t^2} to approximate \int \int\Delta \int\int\Delta=\int\int \theta_0(\omega-\omega_0)\approx \frac{1}{2} (\omega-\omega_0)^2\theta_0(\omega-\omega_0)
We try a different function that approach the \theta, the algebraic function \theta_0^{algebraic}=\frac{c}{2}\left( \frac{x}{\sqrt{1+x^2}}+1\right)\,, \quad x=\frac{\omega_0-\omega}{\sigma}
18.1 Stability
18.2 Z0
\theta_0(\omega-\omega_0) = 0.0593(19) , \chi^2/dof= 0.15622
-\int \theta_0(\omega-\omega_0) = 0.0583(43) , \chi^2/dof= 0.18837
-(\omega-\omega_0)\theta_0(\omega-\omega_0) = 0.0583(43) , \chi^2/dof= 0.18837
0.5(\omega-\omega_0)^2\theta_0(\omega-\omega_0) = 0.0475(80) , \chi^2/dof= 0.20377
\theta_0^{algebraic}(\omega-\omega_0) = 0.0593(19) , \chi^2/dof= 0.1562
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18.3 Z0
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18.4 Z0
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18.5 Z0
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18.6 Z0
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18.7 \sigma extrapolation
18.8 Z0
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