fourier series of $\int_{-\pi}^{\pi}\cos(rx)\cos(kx)dx$

Question

I tried to solve the Fourier series of $$\int_{-\pi}^{\pi}\cos(rx)\cos(kx)dx$$ where $r$ is real number that is not an integer and $k$ is integer. I know that if $r$ and $k$ are real numbers, this Fourier series will be $0$. But how should I solve in this conditions?

Answer 1

$$I =\int^{\pi}_{-\pi}\cos(rx)\cos(kx)dx = 2\int^{\pi}_0\cos(rx)\cos(kx)dx = \int^\pi_0[\cos(r+k)x + \cos(r-k)x\big]dx$$ $$I = \frac{\sin(r+k)x}{r+k} + \frac{\sin(r-k)x}{r-k}\bigg|^\pi_0 = \frac{\sin(r+k)\pi}{r+k} + \frac{\sin(r-k)\pi}{r-k}$$ $$I = \frac{\sin(r\pi)\cos(k\pi) + \cos(r\pi)\sin(k\pi)}{r+k}+ \frac{\sin(r\pi)\cos(k\pi) - \cos(r\pi)\sin(k\pi)}{r-k}$$ $$I = \frac{(-1)^k\sin(r\pi)}{r+k}+\frac{(-1)^k\sin(r\pi)}{r-k} = \frac{2r(-1)^k\sin(r\pi)}{r^2-k^2} , r\in\Bbb{R}-\Bbb{Z},k \in\Bbb{Z}$$