$$$\frac{\sin{\left(\pi \left(z - 1\right) \right)}}{\pi \left(z - 1\right)}$$$ 关于$$$\pi$$$的积分

求$$$\int \frac{\sin{\left(\pi \left(z - 1\right) \right)}}{\pi \left(z - 1\right)}\, d\pi$$$。

对 $$$c=\frac{1}{z - 1}$$$ 和 $$$f{\left(\pi \right)} = \frac{\sin{\left(\pi \left(z - 1\right) \right)}}{\pi}$$$ 应用常数倍法则 $$$\int c f{\left(\pi \right)}\, d\pi = c \int f{\left(\pi \right)}\, d\pi$$$：

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

设$$$u=\pi \left(z - 1\right)$$$。

则$$$du=\left(\pi \left(z - 1\right)\right)^{\prime }d\pi = \left(z - 1\right) d\pi$$$ (步骤见»)，并有$$$d\pi = \frac{du}{z - 1}$$$。

积分变为

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

该积分（正弦积分）没有闭式表达式：

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

回忆一下 $$$u=\pi \left(z - 1\right)$$$:

$$\frac{\operatorname{Si}{\left({\color{red}{u}} \right)}}{z - 1} = \frac{\operatorname{Si}{\left({\color{red}{\pi \left(z - 1\right)}} \right)}}{z - 1}$$

因此，

$$\int{\frac{\sin{\left(\pi \left(z - 1\right) \right)}}{\pi \left(z - 1\right)} d \pi} = \frac{\operatorname{Si}{\left(\pi \left(z - 1\right) \right)}}{z - 1}$$

加上积分常数：

$$\int{\frac{\sin{\left(\pi \left(z - 1\right) \right)}}{\pi \left(z - 1\right)} d \pi} = \frac{\operatorname{Si}{\left(\pi \left(z - 1\right) \right)}}{z - 1}+C$$

$$$\int \frac{\sin{\left(\pi \left(z - 1\right) \right)}}{\pi \left(z - 1\right)}\, d\pi = \frac{\operatorname{Si}{\left(\pi \left(z - 1\right) \right)}}{z - 1} + C$$$A