7.2 Orthogonality of the Trigonometric System
Theorem 7.2.1. The trigonometric system is orthogonal on the interval \(-\pi \leq x \leq \pi \), and hence on any interval of length \(2\pi \): for non-negative integers \(m\) and \(n\), \[\int ^{\pi }_{-\pi } \cos m x \cos n x\,dx = 0\quad ( m\neq n),\qquad \int ^{\pi }_{-\pi }\sin mx \sin nx\,dx = 0 \quad (m\neq n),\] \[\int ^{\pi }_{-\pi }\sin mx \cos nx\,dx = 0\quad (\text {all } m, n) .\] Moreover, for \(m\geq 1\), \(\int ^{\pi }_{-\pi } \cos ^2 mx\,dx = \int ^{\pi }_{-\pi } \sin ^2 mx\,dx = \pi \).
Proof. By the product formulas, \(\cos mx\cos nx = \frac 12[\cos (m - n)x + \cos (m + n)x]\), and for any non-zero integer \(k\), \(\int _{-\pi }^\pi \cos kx\,dx = \left [\frac {\sin kx}{k}\right ]_{-\pi }^\pi = 0\). So for \(m\neq n\) both terms integrate to zero. The same argument with \(\sin mx\sin nx = \frac 12[\cos (m - n)x - \cos (m + n)x]\) gives the second relation, and \(\sin mx\cos nx\) is odd, so its integral over \([-\pi ,\pi ]\) is zero. Finally \(\int ^{\pi }_{-\pi } \cos ^2 mx\,dx = \frac {1}{2}\int ^{\pi }_{-\pi }( 1 + \cos 2mx)\,dx = \pi \), and similarly for \(\sin ^2\). □
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