Decide whether the integral converges. If the integral converges, compute its value.

Question

Quizplus · Accepted Answer

We can use the limit comparison test with $f(x) = \frac{1}{x^2}$ since it also satisfies the conditions of the test: $f(x)$ is positive, continuous, and decreasing for $x \geq 1$.

$$\lim_{x 	o \infty} \frac{\frac{\cos x}{x^2}}{\frac{1}{x^2}} = \lim_{x 	o \infty} \cos x = 	ext{DNE}$$

Since the limit is greater than zero and finite, the integral $\int_1^\infty \frac{\cos x}{x^2} \, dx$ converges if and only if $\int_1^\infty \frac{1}{x^2} \, dx$ converges. But we know that $\int_1^\infty \frac{1}{x^2} \, dx$ does converge, so the given integral also converges. To compute its value, we can use integration by parts with $u = \cos x$ and $dv = \frac{1}{x^2} \, dx$:

$$\int_1^{\infty} \frac{\cos x}{x^2} \, dx = \left[ -\frac{\cos x}{x} ight]_1^\infty + \int_1^{\infty} \frac{\sin x}{x} \, dx$$

The limit as $x 	o \infty$ of $-\frac{\cos x}{x}$ is $0$ since $-1 \leq \cos x \leq 1$ and $\int_1^{\infty} \frac{\sin x}{x} \, dx$ is the well-known Dirichlet integral, which converges to $\frac{\pi}{2}$. Therefore:

$$\int_1^{\infty} \frac{\cos x}{x^2} \, dx = 0 + \frac{\pi}{2} = \frac{\pi}{2}$$

Thus, our answer is D.

Decide Whether the Integral Converges