ベッセル関数の導関数を含む積分公式
不定積分
ベッセル関数の満たす微分方程式を再記して，
\begin{gather}
zu''+u'+\left( \alpha ^2 - \frac{\nu ^2}{z^2} \right) zu = 0 \label{eq:a1}
\end{gather}
式(\ref{eq:a1})\(\times (zu'+u)\)より，
\begin{align}
&\left\{ zu''+u'+ \left( \alpha ^2 - \frac{\nu ^2}{z^2} \right) zu \right\} \big( zu'+u \big) = 0
\nonumber \\
&z^2 u''u' + zu^{\prime 2} + (\alpha^2 z^2 - \nu^2 ) uu'
+ zu''u + u'u + \alpha^2 zu^2 - \frac{\nu^2}{z}u^2
\nonumber \\
&= zu^{\prime 2} - zu^{\prime 2}
\nonumber \\
&\Big( zu^{\prime 2} + z^2u'u'' \Big)
+ \Big\{ \alpha^2 z u^2 + ( \alpha^2 z^2 - \nu^2 ) uu' \Big\}
+ \big( u'u + zu^{\prime 2} + zu''u \big)
\nonumber \\
&= zu^{\prime 2} + \frac{\nu^2}{z}u^2
\nonumber \\
&\frac{d}{dz} \Big( \frac{1}{2} z^2 u^{\prime 2} \Big)
+ \frac{d}{dz} \Big\{ \frac{1}{2} (\alpha^2z^2 - \nu^2 ) u^2 \Big\}
+ \frac{d}{dz} \Big( zuu' \Big)
\nonumber \\
&= z \left( u^{\prime 2} + \frac{\nu^2}{z^2}u^2 \right)
\end{align}
両辺を\(z\) で不定積分して，
\begin{gather}
\int z \left( u^{\prime 2} + \frac{\nu^2}{z^2}u^2 \right) dz
= \frac{z^2}{2} \left\{ u^{\prime 2} + \left( \alpha^2 - \frac{\nu^2}{z^2} \right) u^2
+ \frac{2}{z}u'u \right\}
\end{gather}
これより，次式が得られる．
\begin{eqnarray}
&&\int \left\{ {J_{\nu } '}^2 (\alpha z)
+ \frac{\nu ^2}{\alpha ^2 z^2} J_{\nu }^2(\alpha z) \right\} z \ dz
\nonumber \\
&=& \frac{z^2}{2} \left\{
{J_{\nu }'}^2 (\alpha z)
+ \left( 1- \frac{\nu ^2}{\alpha ^2 z^2} \right) J_{\nu }^2 (\alpha z)
+ \frac{2}{\alpha z} J_{\nu }' (\alpha z) J_{\nu } (\alpha z) \right\}
\end{eqnarray}
また，\(\alpha \neq \beta\) のとき，次のような不定積分公式が得られる（導出省略）．
\begin{eqnarray}
&& \int \left\{ J_{\nu }'(\alpha z) J_{\nu }'(\beta z)
+ \frac{\nu ^2 }{\alpha \beta z^2 } J_{\nu }(\alpha z) J_{\nu }(\beta z) \right\} z \ dz
\nonumber \\
&=& \frac{z}{\alpha ^2 - \beta ^2}
\Big\{ \alpha J_{\nu }(\alpha z) J_{\nu }'(\beta z)
- \beta J_{\nu }'(\alpha z) J_{\nu }(\beta z) \Big\}
\end{eqnarray}
定積分
得られた不定積分の積分範囲を \([0,a]\) とすれば，
\begin{eqnarray}
&& \int _0^a \left\{ J_{\nu }'(\alpha z) J_{\nu }'(\beta z)
+ \frac{\nu ^2 }{\alpha \beta z^2 } J_{\nu }(\alpha z) J_{\nu }(\beta z) \right\} z \ dz
\nonumber \\
&=& \frac{a}{\alpha ^2 - \beta ^2}
\{ \alpha J_{\nu }(\alpha a) J_{\nu }'(\beta z)
- \beta J_{\nu }'(\alpha z) J_{\nu }(\beta z) \} \ \ \ (\alpha \neq \beta )
\\
&& \int _0^a \left\{ {J_{\nu } '}^2 (\alpha z)
+ \frac{\nu ^2}{\alpha ^2 z^2} J_{\nu }^2(\alpha z) \right\} z \ dz
\nonumber \\ %\hspace{5mm}
&=& \frac{a^2}{2} \left\{
{J_{\nu }'}^2 (\alpha a)
+ \left( 1- \frac{\nu ^2}{\alpha ^2 a^2} \right) J_{\nu }^2 (\alpha a)
+ \frac{2}{\alpha a} J_{\nu }' (\alpha a) J_{\nu } (\alpha a) \right\}
\end{eqnarray}
ただし，\(\Re(\nu ) > 0\)．