Evaluation of the singular integral $J$.

Starting from Eq.(), we simplify the denominator in the integral and we introduce the BF index. This gives for the kurtosis

$$\kappa_{40}=6J \times BFI^2,$$

where

$$J={\cal P}\int_{-\infty}^{\infty}\frac{ {\rm d}\nu_1{\rm d}\nu_2{... ...} \frac{e^{-\frac{1}{2}[\nu_1^2+\nu_2^2+\nu_3^2]}} {(\nu_3-\nu_1)(\nu_3-\nu_2)}$$ (54)

Let us determine this integral, which, as will be seen, is not a trivial exercise. Introduce $t_i=nu_i/sqrt{2}$, hence,

$$J=\frac{1}{2\pi^{3/2}}{\cal P}\int_{-\infty}^{\infty} {\rm d}t_1{\rm d}t_2{\rm d}t_3 \frac{e^{-[t_1^2+t_2^2+t_3^2]}} {(t_1-t_3)(t_2-t_3)}$$ (55)

Performing the integration over $t_1$ and $t_2$ first, then

$$J=\frac{1}{2\sqrt{\pi}}{\cal P}\int_{-\infty}^{\infty} {\rm d}t\; Z^2(t)e^{-t^2}$$ (56)

where

$$Z(\zeta)=\frac{1}{\sqrt{\pi}}{\cal P}\int_{-\infty}^{\infty} {\rm d}t\; \frac{e^{-t^2}}{t-\zeta}$$ (57)

is also called the Plasma Dispersion Function. It is customary to express the function $Z$ in terms of the error function with complex argument. However, normally the Residu at $t=zeta$ is included. It is omitted here because only the Principal Value is required. Hence,

$$Z(\zeta)=i\sqrt{\pi}e^{-\zeta^2}\Phi(i\zeta)$$ (58)

where

$$\Phi(i\zeta)=\frac{2}{\sqrt{\pi}}\int_{0}^{i\zeta}{\rm d}t\;e^{-t^2}$$ (59)

is the error function (Note that in the usual Plasma dispersion function an integration from $-infty$ to $0$ is added). Elimination of $Z$ thus gives for $J$

$$J=-\frac{\sqrt{\pi}}{2}{\cal P}\int_{-\infty}^{\infty} {\rm d}t\; e^{-3t^2}\Phi^2(it)$$ (60)

We now evaluate $Phi^2(it)$

$$\Phi^2(it)=\frac{4}{\pi}\int_{0}^{it}{\rm d}x\int_{0}^{it}{\rm d}y \;e^{-\left(x^2+y^2\right)}.$$ (61)

We can perform one integration by introducing polar coordinates

$$x=i\rho\cos \theta,\;y=i\rho\sin \theta$$ (62)

and the result becomes

$$\Phi^2(it)=1-\frac{4}{\pi}\int_{0}^{\pi/4}{\rm d}\theta\; e^{t^2/\cos^2\theta}.$$ (63)

Therefore,

$$J=\frac{2}{\sqrt{\pi}}\int_{0}^{\pi/4}{\rm d}\theta\int_{-\infty}^{\infty} {\rm d}t\;e^{-t^2\left(3-1/\cos^2\theta\right)}-\frac{\pi}{2\sqrt{3}}$$ (64)

Integration over $t$ now gives

$$J=-\frac{\pi}{2\sqrt{3}}+2\int_{0}^{\pi/4}\frac{\cos \theta} {\sqrt{3\cos^2 \theta -1}}.$$ (65)

The remaining integral can be evaluated by means of the transformation $sin theta = z$ and equals $pi/3sqrt{3}$. The final result for $J$ is

$$J=\frac{\pi}{6\sqrt{3}}.$$ (66)