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

$${\cal A}(k) \begin{align*}\equiv 2\,k\,\langle 1\rangle=\frac{2}{\pi}\left\{ \begin{array}{llr} k\,K(k)&~~~~&0\leq k\leq 1\\ [0.5ex] K(1/k)&&k>1\end{array}\right.,\end{align*}$$ (8.112)

$${\cal B}(k) \begin{align*}\equiv \langle \vert X\vert\rangle=\left\{ \begin{array}{llr} (2/\... ...,\sin^{-1}(k)&~~~~&0\leq k\leq 1\\ [0.5ex] 1&&k>1\end{array}\right.,\end{align*}$$ (8.113)

$${\cal C}(k) \begin{align*}\equiv \frac{\langle X^2\rangle}{2\,k}=\frac{2}{\pi}\left\{ \begin... ...,K(k)]/k&~~~~&0\leq k\leq 1\\ [0.5ex] E(1/k)&&k>1\end{array}\right.,\end{align*}$$ (8.114)

$${\cal D}(k) \begin{align*}\equiv \frac{\langle \vert X\vert^3\rangle}{4\,k^2}=\left\{ \begin... ...)]&~~~~&0\leq k\leq 1\\ [0.5ex] 1-1/(2\,k^2)&&k>1\end{array}\right.,\end{align*}$$ (8.115)

$${\cal E}(k) \begin{align*}\equiv \frac{\langle X^4\rangle}{8\,k^3}=\frac{2}{3\pi}\left\{ \be... ...] [2\,(2-1/k^2)\,E(1/k) - (1-1/k^2)\,K(1/k)]&&k>1\end{array}\right.,\end{align*}$$ (8.116)

where

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

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

are standard elliptic integrals [1]. It follows that

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

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

$${\cal I}_1(k) = \frac{4\left[(2\,k^2-1)\,{\cal A} - 2\,k^2\,{\cal C}\right]^2}{{\cal A}},$$ (8.120)

$${\cal I}_2(k) = \left\{\begin{array}{llr} 0 &~~~~& 0\leq k < 1\\ [0.5ex] (1-k\,... ...cal C}/({\cal A}\,{\cal E})\right]/{\cal F}(\infty)&&k\geq 1\end{array}\right.,$$ (8.121)

$${\cal I}_3(k) = \left\{\begin{array}{llr} 0 &~~~~& 0\leq k < 1\\ [0.5ex] k\,{\c... ...C}^2/({\cal A}\,{\cal E})\right]/{\cal F}^2(\infty)&&k\geq 1\end{array}\right.,$$ (8.122)

$${\cal F}(k) =\int_1^k\frac{dk'}{k'^2\,{\cal E}(k')},$$ (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

$$I_1 = 0.8227,$$ (8.124)

$$I_2 = 0.04889,$$ (8.125)

$$I_3 = 0.02944.$$ (8.126)