7–58. Improper integrals Evaluate the following integrals or state that they diverge.
13. ∫ (from 0 to ∞) cos x dx

Verified step by step guidance

1. Recognize that the integral \( \int_0^{\infty} \cos x \, dx \) is an improper integral because the upper limit of integration is infinite.
2. Rewrite the integral as a limit: \( \lim_{t \to \infty} \int_0^t \cos x \, dx \). This allows us to evaluate the integral over a finite interval first and then analyze the behavior as \( t \) approaches infinity.
3. Find the antiderivative of \( \cos x \), which is \( \sin x \). So, \( \int_0^t \cos x \, dx = \sin t - \sin 0 = \sin t \).
4. Evaluate the limit \( \lim_{t \to \infty} \sin t \). Since \( \sin t \) oscillates between -1 and 1 and does not approach a single value, this limit does not exist.
5. Conclude that because the limit does not exist, the improper integral \( \int_0^{\infty} \cos x \, dx \) diverges.

Key Concepts

Improper Integrals

Improper integrals involve integration over an infinite interval or integrands with infinite discontinuities. To evaluate them, limits are used to define the integral as a limit of definite integrals over finite intervals. Determining convergence or divergence is essential.

Convergence and Divergence of Integrals

An integral converges if its limit exists and is finite; otherwise, it diverges. For improper integrals over infinite intervals, convergence depends on the behavior of the integrand as the variable approaches infinity. Testing convergence often involves comparison or limit tests.

Behavior of Oscillatory Functions in Integrals

Functions like cosine oscillate indefinitely without settling to a limit. When integrated over infinite intervals, their oscillations can cause the integral to fail to converge. Understanding how oscillatory behavior affects the integral's limit is crucial for evaluating such integrals.