Answer
Diverges
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^\infty {\cos 2t} dt \cr
& {\text{Using the definition of improper integrals }} \cr
& \int_{ - \infty }^\infty {\cos 2t} dt = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\cos 2t} dt + \mathop {\lim }\limits_{b \to + \infty } \int_0^b {\cos 2t} dt \cr
& {\text{Multiply and divide each integrand by }}\frac{1}{2} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\cos 2t\left( 2 \right)} dt + \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \int_0^b {\cos 2t\left( 2 \right)} dt \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\sin 2t} \right]_a^0 + \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\sin 2t} \right]_0^b \cr
& {\text{Evaluate the integration limits}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\sin 2\left( 0 \right) - \sin 2\left( a \right)} \right] + \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\sin 2\left( b \right) - \sin 2\left( 0 \right)} \right] \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ { - \sin \left( {2a} \right)} \right] + \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\sin \left( {2b} \right)} \right] \cr
& {\text{Evaluate the limits when }}a \to - \infty {\text{ and }}b \to \infty \cr
& = \frac{1}{2}\underbrace {\mathop {\lim }\limits_{a \to - \infty } \left[ { - \sin \left( {2a} \right)} \right]}_{{\text{the limit does not exist}}} + \frac{1}{2}\underbrace {\mathop {\lim }\limits_{b \to + \infty } \left[ {\sin \left( {2b} \right)} \right]}_{{\text{the limit does not exist}}} \cr
& = {\text{the limit does not exist}} \cr
& {\text{Therefore}}{\text{, the integral diverges.}} \cr} $$