Def.1　$x$ の関数 $v_1$, $v_2$, $\cdots$, $v_N$ を縦に並べたもの $\dps{ \vvv=\left( \begin{array}{c} v_1 \\ \vdots \\ v_N \\ \end{array} \right) }$ を $N$ 次元ベクトル値関数と呼ぶ。その微積分は成分ごとに行う： $\dps{ \vvv'=\left( \begin{array}{c} v_1' \\ \vdots \\ v_N' \\ \end{array} \right) }$ , $\qquad \dps{ \int \vvv\,dx=\left( \begin{array}{c} \dps{\int v_1\,dx} \\ \vdots \\ \dps{\int v_N\,dx} \\ \end{array} \right) }$

Ex.2　$\dps{\int e^{ax}\cos(bx)dx}$ を部分積分を使わずに求める方法 $\dps{ \vvv =\left( \begin{array}{c} e^{ax}\cos(bx) \\ e^{ax}\sin(bx) \\ \end{array} \right) }$ とおくと $\dps{ \vvv'=\left( \begin{array}{c} ae^{ax}\cos(bx)-be^{ax}\sin(bx) \\ ae^{ax}\sin(bx)+be^{ax}\cos(bx) \\ \end{array} \right) =\left( \begin{array}{rr} a & -b \\ b & a \\ \end{array} \right) \vvv }$ ∴  $\dps{ \vvv = \left( \begin{array}{rr} a & -b \\ b & a \\ \end{array} \right)^{-1}\vvv' = \frac{1}{a^2+b^2} \left( \begin{array}{rr} a & b \\ -b & a \\ \end{array} \right)\vvv' }$ 両辺積分して $\dps{ \int \vvv\,dx = \frac{1}{a^2+b^2} \left( \begin{array}{rr} a & b \\ -b & a \\ \end{array} \right)\vvv +\left( \begin{array}{c} C \\ D \\ \end{array} \right) }$ この第1成分を見れば $\dps{\int e^{ax}\cos(bx)dx=\frac{1}{a^2+b^2}e^{ax}\big(a\cos(bx)+b\sin(bx)\big)+C}$