Evaluation of Integrals

Figure: 8.1 The integrands ${\cal I}_1(k)$, ${\cal I}_2(k)$, and ${\cal I}_3(k)$.

In order to evaluate the integrals (8.109)–(8.111), it is helpful to define the new magnetic flux-surface label $k=[(1+{\mit\Omega})/2]^{1/2}$. It follows from Equation (8.55) that $k=0$ at the O-points of the magnetic island chain, and $k=1$ on the magnetic separatrix. It is easily demonstrated that

$\displaystyle {\cal A}(k)$ \begin{align*}\equiv 2\,k\,\langle 1\rangle=\frac{2}{\pi}\left\{
\begin{array}{llr} k\,K(k)&~~~~&0\leq k\leq 1\\ [0.5ex]
K(1/k)&&k>1\end{array}\right.,\end{align*} (8.112)
$\displaystyle {\cal B}(k)$ \begin{align*}\equiv \langle \vert X\vert\rangle=\left\{
\begin{array}{llr} (2/\...
...,\sin^{-1}(k)&~~~~&0\leq k\leq 1\\ [0.5ex]
1&&k>1\end{array}\right.,\end{align*} (8.113)
$\displaystyle {\cal C}(k)$ \begin{align*}\equiv \frac{\langle X^2\rangle}{2\,k}=\frac{2}{\pi}\left\{
...,K(k)]/k&~~~~&0\leq k\leq 1\\ [0.5ex]
E(1/k)&&k>1\end{array}\right.,\end{align*} (8.114)
$\displaystyle {\cal D}(k)$ \begin{align*}\equiv \frac{\langle \vert X\vert^3\rangle}{4\,k^2}=\left\{
...)]&~~~~&0\leq k\leq 1\\ [0.5ex]
1-1/(2\,k^2)&&k>1\end{array}\right.,\end{align*} (8.115)
$\displaystyle {\cal E}(k)$ \begin{align*}\equiv \frac{\langle X^4\rangle}{8\,k^3}=\frac{2}{3\pi}\left\{
[2\,(2-1/k^2)\,E(1/k) - (1-1/k^2)\,K(1/k)]&&k>1\end{array}\right.,\end{align*} (8.116)


$\displaystyle E(k)$ $\displaystyle = \int_0^{\pi/2}\left(1-k^2\,\sin^2 u\right)^{1/2}\,du,$ (8.117)
$\displaystyle K(k)$ $\displaystyle = \int_0^{\pi/2}\left(1-k^2\,\sin^2 u\right)^{-1/2}\,du$ (8.118)

are standard elliptic integrals [1]. It follows that

$\displaystyle I_n = \int_0^\infty {\cal I}_n(k)\,dk,$ (8.119)

for $n=1$, $2$, $3$, where

$\displaystyle {\cal I}_1(k)$ $\displaystyle = \frac{4\left[(2\,k^2-1)\,{\cal A} - 2\,k^2\,{\cal C}\right]^2}{{\cal A}},$ (8.120)
$\displaystyle {\cal I}_2(k)$ $\displaystyle = \left\{\begin{array}{llr} 0 &~~~~& 0\leq k < 1\\ [0.5ex]
...cal C}/({\cal A}\,{\cal E})\right]/{\cal F}(\infty)&&k\geq 1\end{array}\right.,$ (8.121)
$\displaystyle {\cal I}_3(k)$ $\displaystyle = \left\{\begin{array}{llr} 0 &~~~~& 0\leq k < 1\\ [0.5ex]
...C}^2/({\cal A}\,{\cal E})\right]/{\cal F}^2(\infty)&&k\geq 1\end{array}\right.,$ (8.122)
$\displaystyle {\cal F}(k)$ $\displaystyle =\int_1^k\frac{dk'}{k'^2\,{\cal E}(k')},$ (8.123)

and use has been made of the easily proved result $d(k\,{\cal C})/dk = {\cal A}$.

The integrands ${\cal I}_1(k)$, ${\cal I}_2(k)$, and ${\cal I}_3(k)$ are shown in Figure 8.1. Note that ${\cal I}_1(k)$ has a logarithmic (and, therefore, integrable) singularity at the island separatrix ($k=1$). The values of the integrals themselves are

$\displaystyle I_1$ $\displaystyle = 0.8227,$ (8.124)
$\displaystyle I_2$ $\displaystyle = 0.04889,$ (8.125)
$\displaystyle I_3$ $\displaystyle = 0.02944.$ (8.126)