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}