第1種と第2種ベッセル関数の不定積分公式
\(A\)，\(B\) を定数として，
\begin{eqnarray}
u &=& A J_{\nu } (\alpha z) + B N_{\nu} (\alpha z)
\\
u' &=& \frac{du}{dz} = \alpha \Big( A J'_\nu (\alpha z) + B N'_\nu (\alpha z) \Big)
\end{eqnarray}
とすると，
\begin{eqnarray}
&&\int z \Big( A J_{\nu } (\alpha z) + B N_\nu (\alpha z) \Big)^2 2dz
\nonumber \\
&=& \frac{1}{2} \left\{ z^2 \Big( A J'_\nu (\alpha z) + B N'_\nu (\alpha z) \Big)^2 \right.
\left. + \left( z^2 - \frac{\nu ^2}{\alpha ^2} \right)
\Big( A J_\nu (\alpha z) + B N_\nu (\alpha z) \Big)^2 \right\}
\end{eqnarray}
上式に\(J_\nu\)，\(N_\nu\) 各々の公式を用いると次式が得られる．
\begin{align}
&\int z J_\nu (\alpha z) N_{\nu } (\alpha z) dz
\nonumber \\
&= \frac{1}{2} \left\{ z^2 J '_\nu (\alpha z) N_{\nu }' (\alpha z)
+ \left( z^2 - \frac{\nu ^2}{\alpha ^2} \right)
J_\nu (\alpha z) N_\nu (\alpha z) \right\} \label{eq:a4-JN}
\end{align}
上の関係式は微分方程式から直接求めることもでき，次のようにすればよい．
\begin{gather}
zu_1''+u_1'+\left( \alpha ^2 - \frac{\nu ^2}{z^2} \right) zu_1 = 0 \label{eq:a1-u1}
\\
zu_2''+u_2'+\left( \alpha ^2 - \frac{\nu ^2}{z^2} \right) zu_2 = 0 \label{eq:a2-u2}
\end{gather}
として\(u_1\)，\(u_2\)を考え，
式\eqref{eq:a1-u1}\(\times u_2'+\)式\eqref{eq:a2-u2}\(\times u_1'\)より，
\begin{align}
&\left\{ zu_1''+u_1'+ \left( \alpha ^2 - \frac{\nu ^2}{z^2} \right) zu_1 \right\} u_2'
+ \left\{ zu_2''+u_2'+ \left( \beta ^2 - \frac{\nu ^2}{z^2} \right) zu_2 \right\} u_1' = 0
\nonumber \\
&2u_1'u_2' + z \big( u_1''u_2'+u_1'u_2'' \big)
+ \left( \alpha ^2 - \frac{\nu ^2}{z^2} \right) z \big( u_1u_2' + u_1'u_2 \big) = 0
\nonumber \\
&2zu_1'u_2' + z^2 \big( u_1''u_2'+u_1'u_2'' \big)
+ \big( \alpha^2 z^2 - \nu^2 \big) \big( u_1u_2' + u_1'u_2 \big) = 0
\nonumber
\end{align}
\begin{gather}
\frac{d}{dz} \Big( z^2u_1'u_2' \big)
+ \big( \alpha^2 z^2 - \nu^2 \big) \frac{d}{dz} \Big( u_1 u_2 \Big) = 0
\end{gather}
さらに，
\begin{align}
&\frac{d}{dz} \Big( z^2u_1' u_2' -\nu^2 u_1u_2 \Big)
+ \alpha^2 z^2 \frac{d}{dz} \Big( u_1 u_2 \Big) = 0
\nonumber \\
&\frac{d}{dz} \Big( z^2u_1' u_2' -\nu^2 u_1u_2 \Big)
+ \alpha ^2 \frac{d}{dz} \Big( z^2 u_1u_2 \Big) = \alpha^2 \cdot 2z u_1 u_2
\end{align}
両辺を\(z\) で不定積分して，
\begin{gather}
z^2u_1' u_2' -\nu^2 u_1u_2+\alpha ^2 z^2 u_1u_2 = 2\alpha^2 \int z u_1 u_2 dz
\end{gather}
よって，\(\alpha \neq 0\)のとき次式となり，式\eqref{eq:a4-JN}が得られる．
\begin{gather}
\int zu_1 u_2 dz
= \frac{1}{2} \left\{ \frac{z^2}{\alpha^2} u_1' u_2'
+ \left( z^2 - \frac{\nu ^2}{\alpha ^2} \right) u_1 u_2 \right\}
\end{gather}