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

$$
\begin{aligned}
I_{1} & = \int_{0}^{\infty} \mathrm{e}^{- \alpha x} \cdot \textcolor{lightgreen}{\cos} \left( \beta x \right) \mathrm{~d} x = ? \\ \\
I_{2} & = \int_{0}^{\infty} \mathrm{e}^{- \alpha x} \cdot \textcolor{pink}{\sin} \left( \beta x \right) \mathrm{~d} x = ?
\end{aligned}
$$

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 $\textcolor{pink}{\sin}$ 与 $\textcolor{lightgreen}{\cos}$ 的性质决定的——

$\textcolor{pink}{\sin}$ 求导会出现 $\textcolor{lightgreen}{\cos}$, 再求导就会重新出现 $\textcolor{pink}{\sin}$;
$\textcolor{lightgreen}{\cos}$ 求导会出现 $\textcolor{pink}{\sin}$, 再求导就会重新出现 $\textcolor{lightgreen}{\cos}$;

$I_{1}$

$$
\begin{aligned}
I_{1} & = \int_{0}^{\infty} \mathrm{e}^{-\alpha x} \textcolor{lightgreen}{\cos} \left( \beta x \right) \mathrm { ~ d } x \\ \\
& = -\frac{1}{\alpha} \int_{0}^{\infty} \textcolor{lightgreen}{\cos} \left( \beta x \right) \mathrm{~d} \left( \mathrm{e}^{-\alpha x} \right) \\ \\
& \Rightarrow \textcolor{gray}{第一次分部积分} \Rightarrow \\ \\
& = – \left. \frac {1}{\alpha} \left[\mathrm{e}^{- \alpha x } \textcolor{lightgreen}{\cos} \left( \beta x \right) \right] \right|_{0}^{\infty} \textcolor{orange}{+} \textcolor{orange}{ \frac{1}{\alpha} } \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \mathrm{~d} \left[ \textcolor{lightgreen}{\cos} \left( \beta x \right) \right] \\ \\
& = \textcolor{violet}{\frac{-1}{\alpha} \left( 0 – 1 \times 1 \right)} \textcolor{red}{-} \textcolor{red}{ \frac{\beta}{\alpha} } \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \textcolor{pink}{\sin} \left( \beta x \right) \mathrm{~d} x \\ \\
& = \textcolor{violet}{\frac{1}{\alpha}} + \frac{\beta}{\alpha^{2}} \int_{0}^{\infty} \textcolor{pink}{\sin} \left( \beta x \right) \cdot \mathrm {~d} \left( \mathrm{e}^{- \alpha x} \right) \\ \\
& \Rightarrow \textcolor{gray}{第二次分部积分} \Rightarrow \\ \\
& = \frac{1}{\alpha} + \left. \frac{\beta}{\alpha^{2}} \left[\mathrm{e}^{ -\alpha x} \textcolor{pink}{\sin} \left( \beta x \right) \right] \right|_{0}^{\infty} – \frac{\beta}{\alpha^{2}} \int_{0}^{\infty} \mathrm{e}^{- \alpha x} \mathrm{~d} \left[ \textcolor{pink}{\sin} \left( \beta x \right) \right] \\ \\
& = \frac{1}{\alpha} + \textcolor{yellow}{ \left. \frac{\beta}{\alpha^{2}} \left[\mathrm{e}^{ -\alpha x} \sin \left( \beta x \right) \right] \right|_{0}^{\infty} } – \frac{\beta^{2}}{\alpha^{2}} \int_{0}^{\infty} \mathrm{e}^{- \alpha x} \textcolor{lightgreen}{\cos} \left( \beta x \right) \mathrm{~d} x \\ \\
& = \frac{1}{\alpha} + \textcolor{yellow}{0} – \frac{\beta^{2}}{\alpha^{2}} \int_{0}^{\infty} \mathrm{e}^{-\alpha x} \textcolor{lightgreen}{\cos} \left( \beta x \right) \mathrm{~d} x \\ \\
& \textcolor{springgreen}{\Large{\boldsymbol{\star}}} \quad \textcolor{springgreen}{\Large{\boldsymbol{\star}}} \quad \textcolor{springgreen}{\Large{\boldsymbol{\star}}} \\ \\
& \Rightarrow \int_{0}^{\infty} \mathrm{e}^{-\alpha x} \textcolor{lightgreen}{\cos} \left( \beta x \right) \mathrm { ~ d } x = \frac{1}{\alpha} – \frac{\beta^{2}}{\alpha^{2}} \int_{0}^{\infty} \mathrm{e}^{-\alpha x} \textcolor{lightgreen}{\cos} \left( \beta x \right) \mathrm{~d} x \\ \\
& \Rightarrow \left(1 + \frac{\beta^{2}}{\alpha^{2}} \right) \int_{0}^{\infty} \mathrm{e}^{-\alpha x} \textcolor{lightgreen}{\cos} \left( \beta x \right) \mathrm{~d} x = \frac{1}{\alpha} \\ \\
& \Rightarrow \int_{0}^{\infty} \mathrm{e}^{-\alpha x } \textcolor{lightgreen}{\cos} \left(\beta x \right) \mathrm{~d} x = \frac{1}{\alpha} \cdot \frac{\alpha^{2}}{\alpha^{2} + \beta^{2}} \\ \\
& \Rightarrow \int_{0}^{\infty} \mathrm{e}^{-\alpha x } \textcolor{lightgreen}{\cos} \left(\beta x \right) \mathrm{~d} x = \textcolor{springgreen}{\boldsymbol{ \frac{\alpha}{\alpha^{2} + \beta^{2}} }}
\end{aligned}
$$

其中，$\alpha > 0$.

$I_{2}$

$$
\begin{aligned}
I_{2} & = \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \textcolor{pink}{\sin} \left( \beta x \right) \mathrm { ~ d } x \\ \\
& = -\frac{1}{\alpha} \int_{0}^{\infty} \textcolor{pink}{\sin} \left( \beta x \right) \mathrm{~d} \left( \mathrm{e}^{ – \alpha x} \right) \\ \\
& \Rightarrow \textcolor{gray}{第一次分部积分} \Rightarrow \\ \\
& = – \left. \frac {1}{\alpha} \left[\mathrm{e}^{- \alpha x } \textcolor{pink}{\sin} \left( \beta x \right) \right] \right|_{0}^{\infty} \textcolor{orange}{+} \textcolor{orange}{ \frac{1}{\alpha} } \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \mathrm{~d} \left[ \textcolor{pink}{\sin} \left( \beta x \right) \right] \\ \\
& = \textcolor{violet}{\frac{-1}{\alpha} \left( 0 – 0 \right)} \textcolor{red}{-} \textcolor{red}{ \frac{\beta}{\alpha} } \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \textcolor{lightgreen}{\cos} \left( \beta x \right) \mathrm{~d} x \\ \\
& = \frac{-\beta}{\alpha^{2}} \int_{0}^{\infty} \textcolor{lightgreen}{\cos} \left( \beta x \right) \cdot \mathrm {~d} \left( \mathrm{e}^{- \alpha x} \right) \\ \\
& \Rightarrow \textcolor{gray}{第二次分部积分} \Rightarrow \\ \\
& = \left. \frac{-\beta}{\alpha^{2}} \left[\mathrm{e}^{ – \alpha x} \textcolor{lightgreen}{\cos} \left( \beta x \right) \right] \right|_{0}^{\infty} + \frac{\beta}{\alpha^{2}} \int_{0}^{\infty} \mathrm{e}^{- \alpha x} \mathrm{~d} \left[ \textcolor{lightgreen}{\cos} \left( \beta x \right) \right] \\ \\
& = \textcolor{yellow}{ \left. \frac{-\beta}{\alpha^{2}} \left[\mathrm{e}^{ – \alpha x} \cos \left( \beta x \right) \right] \right|_{0}^{\infty} } – \frac{\beta^{2}}{\alpha^{2}} \int_{0}^{\infty} \mathrm{e}^{- \alpha x} \textcolor{pink}{\sin} \left( \beta x \right) \mathrm{~d} x \\ \\
& = \textcolor{yellow}{ \frac{- \beta}{\alpha^{2}} \left( 0 – 1 \right) } – \frac{\beta^{2}}{\alpha^{2}} \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \textcolor{pink}{\sin} \left( \beta x \right) \mathrm{~d} x \\ \\
& \textcolor{springgreen}{\Large{\boldsymbol{\star}}} \quad \textcolor{springgreen}{\Large{\boldsymbol{\star}}} \quad \textcolor{springgreen}{\Large{\boldsymbol{\star}}} \\ \\
& \Rightarrow \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \textcolor{pink}{\sin} \left( \beta x \right) \mathrm { ~ d } x = \frac{\beta}{\alpha^{2}} – \frac{\beta^{2}}{\alpha^{2}} \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \textcolor{pink}{\sin} \left( \beta x \right) \mathrm{~d} x \\ \\
& \Rightarrow \left(1 + \frac{\beta^{2}}{\alpha^{2}} \right) \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \textcolor{pink}{\sin} \left( \beta x \right) \mathrm{~d} x = \frac{\beta}{\alpha^{2}} \\ \\
& \Rightarrow \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \textcolor{pink}{\sin} \left(\alpha x \right) \mathrm{~d} x = \frac{\beta}{\alpha^{2}} \cdot \frac{\alpha^{2}}{\alpha^{2} + \beta^{2}} \\ \\
& \Rightarrow \int_{0}^{\infty} \mathrm{e}^{ – \alpha x} \textcolor{pink}{\sin} \left(\alpha x \right) \mathrm{~d} x = \textcolor{springgreen}{\boldsymbol{ \frac{\beta}{\alpha^{2} + \beta^{2}} }}
\end{aligned}
$$

其中，$\alpha > 0$.

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