对含有 $\sin$ 或 $\cos$ 的被积函数做分部积分一般要做两次

一、题目

$$\begin{array}{rl}{I}_1& ={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cdot \cos \left(\beta x\right)\mathrm{d}x=?\\ \\ I_2& ={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cdot \sin \left(\beta x\right)\mathrm{d}x=?\end{array}$$

其中，$\alpha >0$.

二、解析

本文中的两道题目都可以通过使用两次分布积分的方式求解，这也是由三角函数 $\sin$ 与 $\cos$ 的性质决定的——

$\sin$ 求导会出现 $\cos$, 再求导就会重新出现 $\sin$;
$\cos$ 求导会出现 $\sin$, 再求导就会重新出现 $\cos$;

$I_1$

$$\begin{array}{rl}{I}_1& ={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cos \left(\beta x\right)\mathrm{d}x\\ \\ & =-\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}\cos \left(\beta x\right)\mathrm{d}\left({\mathrm{e}}^{-\alpha x}\right)\\ \\ & \Rightarrow \mathrm{第}\mathrm{一}\mathrm{次}\mathrm{分}\mathrm{部}\mathrm{积}\mathrm{分}\Rightarrow \\ \\ & =–{\frac{1}{\alpha }\left[{\mathrm{e}}^{-\alpha x}\cos \left(\beta x\right)\right]|}_0^{\mathrm{\infty }}+\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\mathrm{d}\left[\cos \left(\beta x\right)\right]\\ \\ & =\frac{-1}{\alpha }(0–1\times 1)-\frac{\beta }{\alpha }{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\sin \left(\beta x\right)\mathrm{d}x\\ \\ & =\frac{1}{\alpha }+\frac{\beta }{{\alpha }^2}{\int }_0^{\mathrm{\infty }}\sin \left(\beta x\right)\cdot \mathrm{d}\left({\mathrm{e}}^{-\alpha x}\right)\\ \\ & \Rightarrow \mathrm{第}\mathrm{二}\mathrm{次}\mathrm{分}\mathrm{部}\mathrm{积}\mathrm{分}\Rightarrow \\ \\ & =\frac{1}{\alpha }+{\frac{\beta }{{\alpha }^2}\left[{\mathrm{e}}^{-\alpha x}\sin \left(\beta x\right)\right]|}_0^{\mathrm{\infty }}–\frac{\beta }{{\alpha }^2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\mathrm{d}\left[\sin \left(\beta x\right)\right]\\ \\ & =\frac{1}{\alpha }+{\frac{\beta }{{\alpha }^2}[{\mathrm{e}}^{-\alpha x}\sin \left(\beta x\right)]|}_0^{\mathrm{\infty }}–\frac{{\beta }^2}{{\alpha }^2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cos \left(\beta x\right)\mathrm{d}x\\ \\ & =\frac{1}{\alpha }+0–\frac{{\beta }^2}{{\alpha }^2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cos \left(\beta x\right)\mathrm{d}x\\ \\ & \mathbf{\star }\mathbf{\star }\mathbf{\star }\\ \\ & \Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cos \left(\beta x\right)\mathrm{d}x=\frac{1}{\alpha }–\frac{{\beta }^2}{{\alpha }^2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cos \left(\beta x\right)\mathrm{d}x\\ \\ & \Rightarrow (1+\frac{{\beta }^2}{{\alpha }^2}){\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cos \left(\beta x\right)\mathrm{d}x=\frac{1}{\alpha }\\ \\ & \Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cos \left(\beta x\right)\mathrm{d}x=\frac{1}{\alpha }\cdot \frac{{\alpha }^2}{{\alpha }^2+{\beta }^2}\\ \\ & \Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\cos \left(\beta x\right)\mathrm{d}x=\frac{\mathit{\alpha }}{{\mathit{\alpha }}^{\mathbf{2}}\mathbf{+}{\mathit{\beta }}^{\mathbf{2}}}\end{array}$$

其中，$\alpha >0$.

$I_2$

$$\begin{array}{rl}{I}_2& ={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\sin \left(\beta x\right)\mathrm{d}x\\ \\ & =-\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}\sin \left(\beta x\right)\mathrm{d}\left({\mathrm{e}}^{–\alpha x}\right)\\ \\ & \Rightarrow \mathrm{第}\mathrm{一}\mathrm{次}\mathrm{分}\mathrm{部}\mathrm{积}\mathrm{分}\Rightarrow \\ \\ & =–{\frac{1}{\alpha }\left[{\mathrm{e}}^{-\alpha x}\sin \left(\beta x\right)\right]|}_0^{\mathrm{\infty }}+\frac{1}{\alpha }{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\mathrm{d}\left[\sin \left(\beta x\right)\right]\\ \\ & =\frac{-1}{\alpha }\left(0–0\right)-\frac{\beta }{\alpha }{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\cos \left(\beta x\right)\mathrm{d}x\\ \\ & =\frac{-\beta }{{\alpha }^2}{\int }_0^{\mathrm{\infty }}\cos \left(\beta x\right)\cdot \mathrm{d}\left({\mathrm{e}}^{-\alpha x}\right)\\ \\ & \Rightarrow \mathrm{第}\mathrm{二}\mathrm{次}\mathrm{分}\mathrm{部}\mathrm{积}\mathrm{分}\Rightarrow \\ \\ & ={\frac{-\beta }{{\alpha }^2}\left[{\mathrm{e}}^{–\alpha x}\cos \left(\beta x\right)\right]|}_0^{\mathrm{\infty }}+\frac{\beta }{{\alpha }^2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\mathrm{d}\left[\cos \left(\beta x\right)\right]\\ \\ & ={\frac{-\beta }{{\alpha }^2}[{\mathrm{e}}^{–\alpha x}\cos \left(\beta x\right)]|}_0^{\mathrm{\infty }}–\frac{{\beta }^2}{{\alpha }^2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\alpha x}\sin \left(\beta x\right)\mathrm{d}x\\ \\ & =\frac{-\beta }{{\alpha }^2}\left(0–1\right)–\frac{{\beta }^2}{{\alpha }^2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\sin \left(\beta x\right)\mathrm{d}x\\ \\ & \mathbf{\star }\mathbf{\star }\mathbf{\star }\\ \\ & \Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\sin \left(\beta x\right)\mathrm{d}x=\frac{\beta }{{\alpha }^2}–\frac{{\beta }^2}{{\alpha }^2}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\sin \left(\beta x\right)\mathrm{d}x\\ \\ & \Rightarrow (1+\frac{{\beta }^2}{{\alpha }^2}){\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\sin \left(\beta x\right)\mathrm{d}x=\frac{\beta }{{\alpha }^2}\\ \\ & \Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\sin \left(\alpha x\right)\mathrm{d}x=\frac{\beta }{{\alpha }^2}\cdot \frac{{\alpha }^2}{{\alpha }^2+{\beta }^2}\\ \\ & \Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{–\alpha x}\sin \left(\alpha x\right)\mathrm{d}x=\frac{\mathit{\beta }}{{\mathit{\alpha }}^{\mathbf{2}}\mathbf{+}{\mathit{\beta }}^{\mathbf{2}}}\end{array}$$

其中，$\alpha >0$.

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