Integral of $$$\frac{\sin{\left(\pi \ln\left(x\right) \right)}}{x}$$$

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

Let $$$u=\pi \ln{\left(x \right)}$$$.

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

The integral becomes

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

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

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

The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

Recall that $$$u=\pi \ln{\left(x \right)}$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{\pi} = - \frac{\cos{\left({\color{red}{\pi \ln{\left(x \right)}}} \right)}}{\pi}$$

Therefore,

$$\int{\frac{\sin{\left(\pi \ln{\left(x \right)} \right)}}{x} d x} = - \frac{\cos{\left(\pi \ln{\left(x \right)} \right)}}{\pi}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(\pi \ln{\left(x \right)} \right)}}{x} d x} = - \frac{\cos{\left(\pi \ln{\left(x \right)} \right)}}{\pi}+C$$

$$$\int \frac{\sin{\left(\pi \ln\left(x\right) \right)}}{x}\, dx = - \frac{\cos{\left(\pi \ln\left(x\right) \right)}}{\pi} + C$$$A