Integral of $$$\frac{\sin{\left(\pi \left(z - 1\right) \right)}}{\pi \left(z - 1\right)}$$$ with respect to $$$\pi$$$

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

Apply the constant multiple rule $$$\int c f{\left(\pi \right)}\, d\pi = c \int f{\left(\pi \right)}\, d\pi$$$ with $$$c=\frac{1}{z - 1}$$$ and $$$f{\left(\pi \right)} = \frac{\sin{\left(\pi \left(z - 1\right) \right)}}{\pi}$$$:

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

Let $$$u=\pi \left(z - 1\right)$$$.

Then $$$du=\left(\pi \left(z - 1\right)\right)^{\prime }d\pi = \left(z - 1\right) d\pi$$$ (steps can be seen »), and we have that $$$d\pi = \frac{du}{z - 1}$$$.

Therefore,

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

This integral (Sine Integral) does not have a closed form:

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

Recall that $$$u=\pi \left(z - 1\right)$$$:

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

Therefore,

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

Add the constant of integration:

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

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