$$$\frac{\sin{\left(x \right)}}{x + 1}$$$ 的積分

求$$$\int \frac{\sin{\left(x \right)}}{x + 1}\, dx$$$。

令 $$$u=x + 1$$$。

則 $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

該積分可改寫為

$${\color{red}{\int{\frac{\sin{\left(x \right)}}{x + 1} d x}}} = {\color{red}{\int{\frac{\sin{\left(u - 1 \right)}}{u} d u}}}$$

重寫被積函數:

$${\color{red}{\int{\frac{\sin{\left(u - 1 \right)}}{u} d u}}} = {\color{red}{\int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)} - \sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u}}}$$

Expand the expression:

$${\color{red}{\int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)} - \sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u}}} = {\color{red}{\int{\left(\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} - \frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u}\right)d u}}}$$

逐項積分：

$${\color{red}{\int{\left(\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} - \frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u}\right)d u}}} = {\color{red}{\left(- \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + \int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} d u}\right)}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\cos{\left(1 \right)}$$$ 與 $$$f{\left(u \right)} = \frac{\sin{\left(u \right)}}{u}$$$：

$$- \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + {\color{red}{\int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} d u}}} = - \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + {\color{red}{\cos{\left(1 \right)} \int{\frac{\sin{\left(u \right)}}{u} d u}}}$$

此積分（正弦積分）不存在閉式表示：

$$- \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + \cos{\left(1 \right)} {\color{red}{\int{\frac{\sin{\left(u \right)}}{u} d u}}} = - \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + \cos{\left(1 \right)} {\color{red}{\operatorname{Si}{\left(u \right)}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\sin{\left(1 \right)}$$$ 與 $$$f{\left(u \right)} = \frac{\cos{\left(u \right)}}{u}$$$：

$$\cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - {\color{red}{\int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u}}} = \cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - {\color{red}{\sin{\left(1 \right)} \int{\frac{\cos{\left(u \right)}}{u} d u}}}$$

此積分（餘弦積分）不存在閉式表示：

$$\cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - \sin{\left(1 \right)} {\color{red}{\int{\frac{\cos{\left(u \right)}}{u} d u}}} = \cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - \sin{\left(1 \right)} {\color{red}{\operatorname{Ci}{\left(u \right)}}}$$

回顧一下 $$$u=x + 1$$$：

$$- \sin{\left(1 \right)} \operatorname{Ci}{\left({\color{red}{u}} \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left({\color{red}{u}} \right)} = - \sin{\left(1 \right)} \operatorname{Ci}{\left({\color{red}{\left(x + 1\right)}} \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left({\color{red}{\left(x + 1\right)}} \right)}$$

因此，

$$\int{\frac{\sin{\left(x \right)}}{x + 1} d x} = - \sin{\left(1 \right)} \operatorname{Ci}{\left(x + 1 \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left(x + 1 \right)}$$

加上積分常數：

$$\int{\frac{\sin{\left(x \right)}}{x + 1} d x} = - \sin{\left(1 \right)} \operatorname{Ci}{\left(x + 1 \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left(x + 1 \right)}+C$$

$$$\int \frac{\sin{\left(x \right)}}{x + 1}\, dx = \left(- \sin{\left(1 \right)} \operatorname{Ci}{\left(x + 1 \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left(x + 1 \right)}\right) + C$$$A