3. Buktikan \( \int_{0}^{2 \pi} e^{i m \theta} e^{-i n \theta} d \theta=\left\{\begin{array}{c}0, m \neq n \\ 2 \pi, m=n\end{array}\right. \)

To prove the integral \[ \int_{0}^{2 \pi} e^{i m \theta} e^{-i n \theta} d \theta = \left\{ \begin{array}{c} 0, \quad m \neq n \\ 2 \pi, \quad m = n \end{array} \right. \] we can start by simplifying the integrand. 1. **Combine the Exponents:** The integrand can be rewritten as: \[ e^{i m \theta} e^{-i n \theta} = e^{i (m - n) \theta} \] Thus, the integral becomes: \[ \int_{0}^{2 \pi} e^{i (m - n) \theta} d \theta \] 2. **Evaluate the Integral:** We will evaluate the integral based on the value of \( m - n \). - **Case 1: \( m = n \)** If \( m = n \), then \( m - n = 0 \), and we have: \[ \int_{0}^{2 \pi} e^{0} d \theta = \int_{0}^{2 \pi} 1 \, d \theta = 2 \pi \] - **Case 2: \( m \neq n \)** If \( m \neq n \), then \( m - n \) is a non-zero integer. The integral becomes: \[ \int_{0}^{2 \pi} e^{i (m - n) \theta} d \theta \] The integral of \( e^{i k \theta} \) over one full period (from \( 0 \) to \( 2\pi \)) is zero for any integer \( k \neq 0 \): \[ \int_{0}^{2 \pi} e^{i k \theta} d \theta = 0 \quad \text{for } k \neq 0 \] Therefore, since \( m - n \neq 0 \), we have: \[ \int_{0}^{2 \pi} e^{i (m - n) \theta} d \theta = 0 \] 3. **Conclusion:** Combining both cases, we conclude that: \[ \int_{0}^{2 \pi} e^{i m \theta} e^{-i n \theta} d \theta = \left\{ \begin{array}{c} 0, \quad m \neq n \\ 2 \pi, \quad m = n \end{array} \right. \] This completes the proof.