Pearson from Gaussian

Consider changing $\rho (\sigma )$, the distribution of classes, to

$$\rho _{n}(\sigma )=N\left( \frac{w^{n-1}}{\sigma ^{n}}\right) \exp \left( -% \frac{w^{2}}{2\sigma ^{2}}\right) ,\qquad n>1.$$

The normalization N is

$$N=\frac{1}{2^{\left( n-3\right) /2}\Gamma \left( \frac{n-1}{2}\right) }.$$

Convolution of this with the Gaussian distribution produces Pearson distributions[2]

$$P_{P,b}(B_{i})=\frac{(norm)}{w\left( \frac{B_{i}^{2}}{w^{2}}+1\right) ^{b}}% ,\qquad b=\frac{n}{2}>\frac{1}{2},$$

$$P_{P,b}(\vert{\bf B}\vert)=(norm)\frac{B^{2}}{w^{3}}\left( \frac{1}{\left( \frac{% B^{2}}{w^{2}}+1\right) ^{(b+1)}}\right) .$$

This includes the Lorentzian as the b=1 special case, and the Gaussian as the $b\rightarrow \infty$ limit. While it is thus expected that the relaxation function G_P,b(t) will evolve smoothly from Lorentzian KT to Gaussian KT as b increases, b=2 (n=4) is the only nontrivial case for which closed form G(t) is known:

$$G_{P,2}(t)=\frac{1}{3}+\frac{2}{3}\left( 1+at-a^{2}t^{2}\right) e^{-at}.$$