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

Guida verificata passo dopo passo

Recognize that the integral \( \int_0^{\infty} \cos x \, dx \) is an improper integral because the upper limit of integration is infinite.

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.

Find the antiderivative of \( \cos x \), which is \( \sin x \). So, \( \int_0^t \cos x \, dx = \sin t - \sin 0 = \sin t \).

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.

Conclude that because the limit does not exist, the improper integral \( \int_0^{\infty} \cos x \, dx \) diverges.

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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.

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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.

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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.

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