DTFT of cos ω₀n

The signal $x[n] = \cos\omega_0 n$ is not absolutely summable, so its DTFT contains impulses.

Step 1: Euler's formula

$$\cos\omega_0 n = \frac{1}{2}e^{j\omega_0 n} + \frac{1}{2}e^{-j\omega_0 n}$$

Step 2: DTFT of a complex exponential

$$e^{j\omega_0 n} \xleftrightarrow{DTFT} 2\pi\sum_{l=-\infty}^{\infty}\delta(\omega - \omega_0 - 2\pi l)$$

Check: over one period the inverse DTFT is $\frac{1}{2\pi}\int_{2\pi}2\pi\delta(\omega - \omega_0)e^{j\omega n}d\omega = e^{j\omega_0 n}$. The impulses repeat every $2\pi$ because every DTFT is $2\pi$-periodic.

Step 3: Linearity

$$\boxed{X(e^{j\omega}) = \pi\sum_{l=-\infty}^{\infty}\left[\delta(\omega - \omega_0 - 2\pi l) + \delta(\omega + \omega_0 - 2\pi l)\right]}$$

In the range $-\pi \le \omega < \pi$ this is two impulses of area $\pi$ at $\omega = \pm\omega_0$.

 X(e^jw)
     pi          pi
      ^           ^
      |           |
 --+--+-----+-----+--+--- w
  -pi -w0   0    w0  pi
 (pattern repeats every 2pi)