8.1 調和関数の不定積分
境界条件の異なる２つの固有関数$f_m$（固有値$k_m$），
$g_n$（固有値$\hat{k}_n$）
\begin{gather}
f_m'' + k_m^2 f_m = 0, \ \ \ \ \ g_n'' + \hat{k}_n^2 g_n = 0
\label{eq:fmgn}
\end{gather}
について，
\begin{eqnarray}
\frac{d}{dx} \left( f_m g_n' - f_m' g_n \right)
&=& f_m' g_n' + f_m g_n'' - f_m'' g_n - f_m' g_n'
\nonumber \\
&=& f_m g_n'' - f_m'' g_n
\nonumber \\
&=& f_m (-\hat{k}_n^2 g_n ) - (-k_m^2 f_m) g_n
\nonumber \\
&=& \big( k_m^2 - \hat{k}_n^2 \big) f_m g_n
\nonumber
\end{eqnarray}
不定積分すると，次式が得られる（積分定数省略）．
\begin{gather}
f_m g_n' - f_m' g_n
= \big( k_m^2 - \hat{k}_n^2 \big) \int f_m g_n dx
\end{gather}
よって，$k_m \neq \hat{k}_{n}$のとき，調和関数に関する不定積分は次のようになる．
\begin{gather}
\int f_m g_n dx = \frac{f_m g_n' - f_m' g_n}{k_m^2 - \hat{k}_n^2 }
\label{eq:int_fmgn}
\end{gather}
正弦・正弦関数（$\sin$，$\sin$）の不定積分
まず，$f$，$g$をともに正弦関数
\begin{gather}
f = \sin (k_x x + \xi ), \ \ \ \ \
g = \sin (\hat{k}_x x + \hat{\xi})
\end{gather}
とする．これらを$x$で微分すると，
\begin{gather}
f' = k_x \cos (k_x x + \xi), \ \ \ \
g' = \hat{k}_x \cos (\hat{k}_x x + \hat{\xi})
\end{gather}
先に求めた不定積分の公式\eqref{eq:int_fmgn}より，
\begin{eqnarray}
&&\int f g dx
\nonumber \\
&=& \int \sin (k_x x + \xi) \cdot \sin (\hat{k}_x x + \hat{\xi}) dx
\nonumber \\
&=& \frac{\sin (k_x x + \xi ) \cdot \hat{k}_x \cos (\hat{k}_x x + \hat{\xi})
- k_x \cos (k_x x + \xi) \cdot \sin (\hat{k}_x x + \hat{\xi})}{k_x^2 - \hat{k}_x^2}
\end{eqnarray}
三角関数の積和公式
\begin{align}
&2 \sin \alpha \cos \beta = \sin (\alpha + \beta) + \sin (\alpha - \beta)
\\
&2 \cos \alpha \sin \beta = \sin (\alpha + \beta) - \sin (\alpha - \beta)
\end{align}
より，
\begin{eqnarray}
&&\int f g dx
\nonumber \\
&=& \hat{k}_x \frac{\sin \big\{ (k_x+\hat{k}_x)x + (\xi+\hat{\xi}) \big\}
+ \sin \big\{ (k_x-\hat{k}_x)x + (\xi-\hat{\xi}) \big\} }{2(k_x^2 - \hat{k}_x^2)}
\nonumber \\
&&- k_x \frac{\sin \big\{ (k_x+\hat{k}_x)x + (\xi+\hat{\xi}) \big\}
- \sin \big\{ (k_x-\hat{k}_x)x + (\xi-\hat{\xi}) \big\} }{2(k_x^2 - \hat{k}_x^2)}
\nonumber \\
&=& -\frac{\sin \big\{ (k_x+\hat{k}_x)x + (\xi+\hat{\xi}) \big\} }{2(k_x+\hat{k}_x)}
+ \frac{\sin \big\{ (k_x-\hat{k}_x)x + (\xi-\hat{\xi}) \big\} }{2(k_x-\hat{k}_x)}
\end{eqnarray}
ここで，加法定理より，
\begin{align}
&\sin \big\{ (k_x-\hat{k}_x)x + (\xi-\hat{\xi}) \big\}
\nonumber \\
&= \sin \big\{ (k_x-\hat{k}_x)x \big\} \cos (\xi-\hat{\xi})
+ \cos \big\{ (k_x-\hat{k}_x)x \big\} \sin (\xi-\hat{\xi})
\end{align}
$\hat{k}_x \to k_x$のとき，
\begin{gather}
\lim_{\hat{k}_x \to k_x} \frac{\sin \big\{ (k_x-\hat{k}_x)x \big\}
\cos (\xi-\hat{\xi})}{2(k_x-\hat{k}_x)}
= \frac{x}{2} \cos (\xi-\hat{\xi})
\end{gather}
$k_x = \hat{k}_x$のとき，
\begin{gather}
\int f g dx
= -\frac{\sin \big\{ 2k_x x + (\xi+\hat{\xi}) \big\} }{4 k_x} + \frac{x \cos (\xi-\hat{\xi})}{2}
\end{gather}
また，加法定理より，
\begin{gather}
\sin \big\{ 2k_x x + (\xi+\hat{\xi}) \big\}
= \sin (2k_x x) \cos (\xi+\hat{\xi}) + \cos (2k_x x) \sin (\xi+\hat{\xi})
\end{gather}
$k_x \to 0$ のとき，
\begin{gather}
\lim_{k_x \to 0} \frac{\sin (2k_x x) \cos (\xi+\hat{\xi})}{4k_x}
= \frac{x}{2} \cos (\xi+\hat{\xi})
\end{gather}
$k_x = 0$ のとき，三角関数の和積公式
\begin{gather}
-\cos A + \cos B = 2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}
\end{gather}
より，
\begin{eqnarray}
\int f g dx
&=& -\frac{x \cos (\xi+\hat{\xi})}{2}+ \frac{x \cos (\xi-\hat{\xi})}{2}
\nonumber \\
&=& x \sin \xi \sin\hat{\xi}
\end{eqnarray}
あるいは，$k_x = \hat{k}_x = 0$ のとき，
$f = \sin \xi$，$g = \sin \hat{\xi}$より，
\begin{eqnarray}
\int f g dx
&=& \sin \xi \sin \hat{\xi} \int dx
\nonumber \\
&=& x \sin \xi \sin \hat{\xi}
\end{eqnarray}
また，$k_x = \hat{k}_x \ne 0$のとき，
$f = \sin (k_x x + \xi )$，
$g = \sin (k_x x + \hat{\xi})$．
三角関数の積和公式
\begin{gather}
2 \sin \alpha \sin \beta = -\cos (\alpha + \beta) + \cos (\alpha - \beta)
\end{gather}
より，
\begin{eqnarray}
fg &=& \sin (k_x x + \xi ) \cdot \sin (k_x x + \hat{\xi})
\nonumber \\
&=& \frac{1}{2} \Big[ -\cos \big\{ 2k_x x + (\xi + \hat{\xi}) \big\} + \cos (\xi - \hat{\xi}) \Big]
\end{eqnarray}
よって，
\begin{eqnarray}
\int f g dx
&=& \frac{1}{2} \int \Big[ -\cos \big\{ 2k_x x + (\xi + \hat{\xi}) \big\} + \cos (\xi - \hat{\xi}) \Big] dx
\nonumber \\
&=& -\frac{\sin \big\{ 2k_x x + (\xi+\hat{\xi}) \big\} }{4 k_x} + \frac{x \cos (\xi-\hat{\xi})}{2}
\end{eqnarray}
正弦・正弦関数の積分をまとめると，$k_x \ne \hat{k}_x$のとき，
\begin{align}
&\int \sin (k_x x + \xi) \cdot \sin (\hat{k}_x x + \hat{\xi}) dx
\nonumber \\
&= -\frac{\sin \big\{ (k_x+\hat{k}_x)x + (\xi+\hat{\xi}) \big\} }{2(k_x+\hat{k}_x)}
+ \frac{\sin \big\{ (k_x-\hat{k}_x)x + (\xi-\hat{\xi}) \big\} }{2(k_x-\hat{k}_x)}
\end{align}
また，$k_x = \hat{k}_x \ne 0$ のとき，
\begin{align}
&\int \sin (k_x x + \xi) \cdot \sin (k_x x + \hat{\xi}) dx
\nonumber \\
&= -\frac{\sin \big\{ 2k_x x + (\xi+\hat{\xi}) \big\} }{4 k_x} + \frac{x \cos (\xi-\hat{\xi})}{2}
\end{align}
さらに，$k_x = \hat{k}_x = 0$のとき，
\begin{gather}
%\int f g dx =
\int \sin \xi \sin \hat{\xi} dx = x \sin \xi \sin \hat{\xi}
\end{gather}
余弦・余弦関数（$\cos$，$\cos$）の不定積分
次に，$f$，$g$をともに余弦関数
\begin{gather}
f = \cos (k_x x + \xi ), \ \ \ \ \
g = \cos (\hat{k}_x x + \hat{\xi})
\end{gather}
とする．ここで，$\xi \equiv \zeta + \pi/2$，$\hat{\xi} \equiv \hat{\zeta} + \pi/2$
とおくと，
\begin{eqnarray}
f &=& \cos (k_x x + \zeta + \pi/2 ) = \sin (k_x x + \zeta )
\\
g &=& \cos (\hat{k}_x x + \hat{\zeta} + \pi/2) = \sin (\hat{k}_x x + \hat{\zeta} )
\end{eqnarray}
ここで，$\zeta = \xi - \pi/2$，$\hat{\zeta} = \hat{\xi} - \pi/2$より，
\begin{align}
&\zeta + \hat{\zeta} = \xi + \hat{\xi} - \pi
\\
&\zeta - \hat{\zeta} = \xi - \hat{\xi}
\end{align}
よって，
\begin{eqnarray}
&&\int f g dx
\nonumber \\
&=& \int \cos (k_x x + \xi) \cdot \cos (\hat{k}_x x + \hat{\xi}) dx
\nonumber \\
&=& \int \sin (k_x x + \zeta) \cdot \sin (\hat{k}_x x + \hat{\zeta}) dx
\nonumber \\
&=& -\frac{\sin \big\{ (k_x+\hat{k}_x)x + (\zeta+\hat{\zeta}) \big\} }{2(k_x+\hat{k}_x)}
+ \frac{\sin \big\{ (k_x-\hat{k}_x)x + (\zeta-\hat{\zeta}) \big\} }{2(k_x-\hat{k}_x)}
\nonumber \\
&=& \frac{\sin \big\{ (k_x+\hat{k}_x)x + (\xi+\hat{\xi}) \big\} }{2(k_x+\hat{k}_x)}
+ \frac{\sin \big\{ (k_x-\hat{k}_x)x + (\xi-\hat{\xi}) \big\} }{2(k_x-\hat{k}_x)}
\end{eqnarray}
また，$k_x = \hat{k}_x \ne 0$のとき，
\begin{eqnarray}
\int f g dx
&=& \int \cos (k_x x + \xi) \cdot \cos (k_x x + \hat{\xi}) dx
\nonumber \\
&=& \int \sin (k_x x + \zeta) \cdot \sin (k_x x + \hat{\zeta}) dx
\nonumber \\
&=& -\frac{\sin \big\{ 2k_x x + (\zeta+\hat{\zeta}) \big\} }{4 k_x} + \frac{x \cos (\zeta-\hat{\zeta})}{2}
\nonumber \\
&=& \frac{\sin \big\{ 2k_x x + (\xi+\hat{\xi}) \big\} }{4 k_x} + \frac{x \cos (\xi-\hat{\xi})}{2}
\end{eqnarray}
さらに，$k_x = \hat{k}_x = 0$のとき，
\begin{eqnarray}
\int f g dx
&=& \int \cos \xi \cos \hat{\xi} dx
\nonumber \\
&=& x \cos \xi \cos \hat{\xi}
\end{eqnarray}
不定積分のまとめ
正弦・正弦関数を上側，余弦・余弦関数を下側に記して
積分をまとめると，$k_x \ne \hat{k}_x$ のとき，
\begin{eqnarray}
&&\int \begin{matrix} \sin \\ \cos \end{matrix} (k_x x + \xi)
\cdot \begin{matrix} \sin \\ \cos \end{matrix} (\hat{k}_x x + \hat{\xi}) dx
\nonumber \\
&=& \mp \frac{\sin \big\{ (k_x+\hat{k}_x)x + (\xi+\hat{\xi}) \big\}}{2(k_x+\hat{k}_x)}
+ \frac{\sin \big\{ (k_x-\hat{k}_x)x + (\xi-\hat{\xi}) \big\} }{2(k_x-\hat{k}_x)}
\end{eqnarray}
また，$k_x = \hat{k}_x \ne 0$ のとき，
\begin{align}
&\int \begin{matrix} \sin \\ \cos \end{matrix} (k_x x + \xi)
\cdot \begin{matrix} \sin \\ \cos \end{matrix} (k_x x + \hat{\xi}) dx
\nonumber \\
&= \mp \frac{\sin \big\{ 2k_x x + (\xi+\hat{\xi}) \big\} }{4 k_x} + \frac{x \cos (\xi-\hat{\xi})}{2}
\end{align}
さらに，$k_x = \hat{k}_x = 0$ のとき，
\begin{gather}
%\int f g dx =
\int \begin{matrix} \sin \\ \cos \end{matrix} \ \xi
\cdot \begin{matrix} \sin \\ \cos \end{matrix} \ \hat{\xi} dx
= x \ \begin{matrix} \sin \\ \cos \end{matrix} \ \xi
\cdot \begin{matrix} \sin \\ \cos \end{matrix} \ \hat{\xi}
\end{gather}
被積分関数の定数項を若干，変形した形について，$k_x \ne \hat{k}_x$ のとき，
\begin{eqnarray}
&&\int \begin{matrix} \sin \\ \cos \end{matrix} \big\{ k_x (x +x_1) \big\}
\cdot \begin{matrix} \sin \\ \cos \end{matrix} \big\{ \hat{k}_x (x +x_2) \big\} dx
\nonumber \\
&=& \mp \frac{\sin \big\{ k_x (x +x_1) + \hat{k}_x (x +x_2) \big\} }{2(k_x+\hat{k}_x)}
\nonumber \\
&&+ \frac{\sin \big\{ k_x (x +x_1) - \hat{k}_x (x +x_2) \big\} }{2(k_x-\hat{k}_x)}
\end{eqnarray}
また，$k_x = \hat{k}_x \ne 0$ のとき，
\begin{align}
&\int \begin{matrix} \sin \\ \cos \end{matrix} \big\{ k_x (x +x_1) \big\}
\cdot \begin{matrix} \sin \\ \cos \end{matrix} \big\{ k_x (x +x_2) \big\} dx
\nonumber \\
&= \mp \frac{\sin \big\{ k_x (2x + x_1 + x_2) \big\} }{4 k_x}
+ \frac{x \cos \big\{ k_x (x_1-x_2) \big\} }{2}
\end{align}
さらに，$k_x = \hat{k}_x = 0$ のとき，
\begin{gather}
\int \begin{matrix} \sin \\ \cos \end{matrix} \big\{ k_x (x +x_1) \big\}
\cdot \begin{matrix} \sin \\ \cos \end{matrix} \big\{ k_x (x +x_2) \big\} dx
= \begin{matrix} 0 \\ x \end{matrix}
\end{gather}