フラウンホーファ領域
円形導波管モードによる放射界を計算するため，
\begin{eqnarray}
\bar{N}_x
&=& a^2 \frac{A_{mn} \bar{\chi}_{mn}}{2a} \int_0^{2\pi} \int_0^1
\Big\{ J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \cos \\ \sin \end{matrix} (m-1) \phi'
\nonumber \\
&&+ \ell J_{m+1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \cos \\ \sin \end{matrix} (m+1) \phi' \Big\}
e^{ju \bar{\rho} \cos (\phi - \phi')} \bar{\rho} d\bar{\rho} d\phi'
\\
\bar{N}_y
&=& a^2 \frac{A_{mn} \bar{\chi}_{mn}}{2a} \int_0^{2\pi} \int_0^1
\Big\{ -J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \sin \\ -\cos \end{matrix} (m-1) \phi'
\nonumber \\
&&+ \ell J_{m+1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \sin \\ -\cos \end{matrix} (m+1) \phi' \Big\}
e^{ju \bar{\rho} \cos (\phi - \phi')} \bar{\rho} d\bar{\rho} d\phi'
\end{eqnarray}
における積分
\begin{eqnarray}
I_{Nx1}
&=& \int_0^{2\pi} \int_0^1
J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \cos \\ \sin \end{matrix} (m-1) \phi'
e^{ju \bar{\rho} \cos (\phi - \phi')} \bar{\rho} d\bar{\rho} d\phi'
\\
I_{Nx2}
&=& \int_0^{2\pi} \int_0^1
J_{m+1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \cos \\ \sin \end{matrix} (m+1) \phi'
e^{ju \bar{\rho} \cos (\phi - \phi')} \bar{\rho} d\bar{\rho} d\phi'
\\
I_{Ny1}
&=& \int_0^{2\pi} \int_0^1
-J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \sin \\ -\cos \end{matrix} (m-1) \phi'
e^{ju \bar{\rho} \cos (\phi - \phi')} \bar{\rho} d\bar{\rho} d\phi'
\\
I_{Ny2}
&=& \int_0^{2\pi} \int_0^1
J_{m+1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \sin \\ -\cos \end{matrix} (m+1) \phi'
e^{ju \bar{\rho} \cos (\phi - \phi')} \bar{\rho} d\bar{\rho} d\phi'
\end{eqnarray}
を，ベッセル - フーリエ級数（Bessel-Fourier series）
\begin{gather}
e^{j\lambda \rho \cos (\phi - \phi')}
= J_0 (\lambda \rho) + \sum_{n=1}^\infty 2j^n J_n(\lambda \rho ) \cos n(\phi - \phi')
\end{gather}
を用いて積分する．
まず，$\bar{N}_x$の第1項の積分$I_{Nx1}$は，
\begin{eqnarray}
I_{Nx1}
&=& \int_0^{2\pi} \hspace{-2mm} \int_0^1
J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \cos \\ \sin \end{matrix} (m-1) \phi'
\nonumber \\
&&\cdot \left\{ J_0 (u \bar{\rho}) + \sum_{n'=1}^\infty
2j^{n'} J_{n'} (u \bar{\rho} ) \cos n' (\phi - \phi') \right\}
\bar{\rho} d\bar{\rho} d\phi'
\nonumber \\
&=& \int_0^1 J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right) J_0 (u \bar{\rho})
\bar{\rho} d\bar{\rho}
\int _0^{2\pi} \begin{matrix} \cos \\ \sin \end{matrix} (m-1) \phi' d\phi'
\nonumber \\
&&+ \sum_{n'=1}^\infty 2j^{n'}
\int_0^1 J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right) J_{n'} (u \bar{\rho})
\bar{\rho} d\bar{\rho}
\nonumber \\
&&\cdot \int _0^{2\pi} \begin{matrix} \cos \\ \sin \end{matrix} (m-1) \phi' \cos n' (\phi - \phi') d\phi'
\end{eqnarray}
ここで，
\begin{align}
&\int _0^{2\pi} \cos (m \mp 1) \phi' d\phi' = 2\pi \delta _{m,\pm 1}
\\
&\int _0^{2\pi} \sin (m \mp 1) \phi' d\phi' = 0
\end{align}
また，
\begin{eqnarray}
&&\cos (m \mp 1) \phi' \cos n' (\phi - \phi')
\nonumber \\
&=& \frac{1}{2} \Big[ \cos \Big( (m \mp 1) \phi' + n' (\phi - \phi') \Big)
+ \cos \Big( (m \mp 1) \phi' - n' (\phi - \phi') \Big) \Big]
\nonumber \\
&=& \frac{1}{2} \Big[ \cos \Big( (m \mp 1-n') \phi' + n' \phi \Big)
+ \cos \Big( (m \mp 1+n') \phi' - n' \phi \Big) \Big]
\nonumber \\
&=& \frac{1}{2} \Big[ \cos (m \mp 1-n') \phi' \cos n' \phi - \sin (m \mp 1-n') \phi' \sin n' \phi
\nonumber \\
&&+ \cos (m \mp 1+n') \phi' \cos n' \phi + \sin (m \mp 1+n') \phi' \sin n' \phi \Big]
\end{eqnarray}
これを積分すると，$1 \leq n'=m \mp 1$，$-m \pm 1$（$m=0,1,2, \cdots$）のとき値をもち，このとき，
\begin{eqnarray}
\int _0^{2\pi} \cos (m \mp 1) \phi' \cos n' (\phi - \phi') d\phi'
&=& \frac{1}{2} \cos n' \phi \int _0^{2\pi} d\phi'
\nonumber \\
&=& \pi \cos n' \phi
\end{eqnarray}
上側符号では，$m=0$ のとき $n' = -m+1 = 1$，
$m=1$ のとき $n' \geq 1$ となるケースはなし，
$m=2$ のとき $n' = m-1 = 1$，
$m=3,4, \cdots$ のとき $n'=m-1$ である．したがって，
\begin{align}
&\int _0^{2\pi} \cos (m-1) \phi' \cos n' (\phi - \phi') d\phi'
\nonumber \\
&= \left\{
\begin{array}{cl}
\pi \cos |m-1| \phi & (n' = |m-1|, \ m=0,2,3,4,\cdots) \\
0 & (\mbox{otherwise})
\end{array} \right.
\end{align}
また，下側符号では，$m=0$ のとき $n' = m+1 = 1$，
$m=1,2,\cdots $ のとき $n'=m+1$ である．したがって，
\begin{align}
&\int _0^{2\pi} \cos (m+1) \phi' \cos n' (\phi - \phi') d\phi'
\nonumber \\
&= \left\{
\begin{array}{cl}
\pi \cos (m+1) \phi & (n' = m+1) \\
0 & (\mbox{otherwise})
\end{array} \right.
\end{align}
同様にして（導出省略），
\begin{align}
&\int _0^{2\pi} \sin (m-1) \phi' \cos n' (\phi - \phi') d\phi'
\nonumber \\
&= \left\{
\begin{array}{cl}
\pi \sin |m-1| \phi & (n' = |m-1|, \ m=0,2,3,4,\cdots) \\
0 & (\mbox{otherwise})
\end{array} \right.
\\
&\int _0^{2\pi} \sin (m+1) \phi' \cos n' (\phi - \phi') d\phi'
\nonumber \\
&= \left\{
\begin{array}{cl}
\pi \sin (m+1) \phi & (n' = m+1) \\
0 & (\mbox{otherwise})
\end{array} \right.
\end{align}
よって，積分項は，
\begin{gather}
I_{Nx1}
= j^{m-1} 2\pi \ \begin{matrix} \cos \\ \sin \end{matrix} (m-1) \phi
\int_0^1 J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right) J_{m-1} (u \bar{\rho}) \bar{\rho} d\bar{\rho}
\end{gather}
同様にして，$\bar{N}_x$の第2項の積分$I_{Nx2}$は，
\begin{eqnarray}
I_{Nx2}
&=& \int_0^{2\pi} \hspace{-2mm} \int_0^1
J_{m+1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \cos \\ \sin \end{matrix} (m+1) \phi'
\nonumber \\
&&\cdot \left\{ J_0 (u \bar{\rho}) + \sum_{n'=1}^\infty
2j^{n'} J_{n'} (u \bar{\rho} ) \cos n' (\phi - \phi') \right\}
\bar{\rho} d\bar{\rho} d\phi'
\nonumber \\
&=& j^{m+1} 2\pi \ \begin{matrix} \cos \\ \sin \end{matrix} (m+1) \phi
\int_0^1 J_{m+1} \left( \bar{\chi}_{mn} \bar{\rho} \right) J_{m+1} (u \bar{\rho}) \bar{\rho} d\bar{\rho}
\end{eqnarray}
さらに，$\bar{N}_y$の第1項の積分$I_{Ny1}$は，
\begin{eqnarray}
I_{Ny1}
&=& \int_0^{2\pi} \hspace{-2mm}
\int_0^1 -J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \sin \\ -\cos \end{matrix} (m-1) \phi'
\cdot e^{ju \bar{\rho} \cos (\phi - \phi')} \bar{\rho} d\bar{\rho} d\phi'
\nonumber \\
&=& -j^{m-1} 2\pi \ \begin{matrix} \sin \\ -\cos \end{matrix} (m-1) \phi
\int_0^1 J_{m-1} \left( \bar{\chi}_{mn} \bar{\rho} \right) J_{m-1} (u \bar{\rho}) \bar{\rho} d\bar{\rho}
\end{eqnarray}
また，$\bar{N}_y$の第2項の積分$I_{Ny2}$は，
\begin{eqnarray}
I_{Ny2}
&=& \int_0^{2\pi} \hspace{-2mm} \int_0^1
J_{m+1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
\begin{matrix} \sin \\ -\cos \end{matrix} (m+1) \phi'
\cdot e^{ju \bar{\rho} \cos (\phi - \phi')} \bar{\rho} d\bar{\rho} d\phi'
\nonumber \\
&=& j^{m+1} 2\pi \ \begin{matrix} \sin \\ -\cos \end{matrix} (m+1) \phi
\int_0^1 J_{m+1} \left( \bar{\chi}_{mn} \bar{\rho} \right) J_{m+1} (u \bar{\rho}) \bar{\rho} d\bar{\rho}
\end{eqnarray}
積分項は，
\begin{gather}
I_{m \pm 1,n} \equiv
\int_0^1 J_{m \pm 1} \left( \bar{\chi}_{mn} \bar{\rho} \right)
J_{m \pm 1} (u \bar{\rho}) \bar{\rho} d\bar{\rho}
\end{gather}
ベッセル関数の不定積分公式 $ (\alpha \neq \beta )$
\begin{gather}
\int z J_{\nu }(\alpha z) J_{\nu }(\beta z) dz
= \frac{z}{\alpha ^2 - \beta ^2}
\left\{ \beta J_{\nu }(\alpha z) J_{\nu }'(\beta z)
- \alpha J_{\nu }'(\alpha z) J_{\nu }(\beta z) \right\}
\end{gather}
より，
\begin{gather}
\int_0^1 z J_{\nu }(\alpha z) J_{\nu }(\beta z) dz
= \frac{1}{\alpha ^2 - \beta ^2}
\left\{ \beta J_{\nu }(\alpha) J_{\nu }'(\beta)
- \alpha J_{\nu }'(\alpha) J_{\nu }(\beta) \right\}
\end{gather}
よって，
\begin{gather}
I_{m \pm 1,n}
= \frac{1}{\bar{\chi}_{mn}^2 - u^2}
\left\{ u J_{m \pm 1}(\bar{\chi}_{mn}) J_{m \pm 1}'(u)
- \bar{\chi}_{mn} J_{m \pm 1}'(\bar{\chi}_{mn}) J_{m \pm 1}(u) \right\}
\nonumber
\end{gather}