Integral of $$$\frac{k \sin{\left(\frac{x}{k} \right)}}{x}$$$ with respect to $$$x$$$

Find $$$\int \frac{k \sin{\left(\frac{x}{k} \right)}}{x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=k$$$ and $$$f{\left(x \right)} = \frac{\sin{\left(\frac{x}{k} \right)}}{x}$$$:

$${\color{red}{\int{\frac{k \sin{\left(\frac{x}{k} \right)}}{x} d x}}} = {\color{red}{k \int{\frac{\sin{\left(\frac{x}{k} \right)}}{x} d x}}}$$

Let $$$u=\frac{x}{k}$$$.

Then $$$du=\left(\frac{x}{k}\right)^{\prime }dx = \frac{dx}{k}$$$ (steps can be seen »), and we have that $$$dx = k du$$$.

The integral becomes

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

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

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

Recall that $$$u=\frac{x}{k}$$$:

$$k \operatorname{Si}{\left({\color{red}{u}} \right)} = k \operatorname{Si}{\left({\color{red}{\frac{x}{k}}} \right)}$$

Therefore,

$$\int{\frac{k \sin{\left(\frac{x}{k} \right)}}{x} d x} = k \operatorname{Si}{\left(\frac{x}{k} \right)}$$

Add the constant of integration:

$$\int{\frac{k \sin{\left(\frac{x}{k} \right)}}{x} d x} = k \operatorname{Si}{\left(\frac{x}{k} \right)}+C$$

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