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How to prove this equality below

$$\int_{a}^{b}f\left ( x \right )\cot\left ( x \right )\mathrm{d}x=2\sum_{n=1}^{\infty }\int_{a}^{b}f\left ( x \right )\sin\left ( 2nx \right )\mathrm{d}x$$

The series RHS seems to be divergence...I don't know how to do first.

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Hint:

$$2\sum_{n=1}^{\infty }\int_{a}^{b}f\left ( x \right )\sin\left ( 2nx \right )\mathrm{d}x=\int_{a}^{b}f(x)\left[2\sum_{n=1}^{\infty}\sin\left ( 2nx \right )\right]dx$$

So essentially you have to show that

$$\cot(x)=2\sum_{n=1}^{\infty}\sin(2nx)$$

Use Eulers formula $e^{i2nx}=\cos(2nx)+i\sin(2nx)$ to show this. E.g. sum over $e^{i2nx}$ and then take the maginary part of the result.

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