Question

Determine whether the following integrals converge or diverge. $$\int_{1}^{\infty} \frac{2+\cos x}{x^{2}} d x$$

Short answer

Question: Determine if the following integral converges or diverges: $$\int_{1}^{\infty} \frac{2+\cos x}{x^{2}} d x$$ Answer: The integral converges.

Step-by-step solution

Comparison function

We're going to compare our given function with \(\frac{1}{x^2}\). Notice that for our function: $$\frac{2+\cos x}{x^2} \geq \frac{1}{x^2} $$ as \(\cos x\) is always greater than or equal to \(-1\). Now let's find the integral of the comparison function with respect to \(x\).

Integrate the comparison function

Integrate \(\frac{1}{x^2}\) with respect to \(x\) from \(1\) to \(\infty\): $$\int_{1}^{\infty} \frac{1}{x^2} d x$$ Let's use the substitution method for this integral. Let \(u=x^2\), then \(\frac{du}{dx} = 2x\) or \(dx=\frac{du}{2x}\). When \(x=1\), \(u=1\). When \(x \to \infty\), \(u \to \infty\). Now substitute \(u\) and \(dx\) into the integral and then solve, we get: $$\int_{1}^{\infty} \frac{1}{x^2} d x = \int_1^\infty \frac{du}{2u} = \frac{1}{2}\int_1^\infty u^{-1} du$$ Now integrate \(u^{-1}\) and evaluate the improper integral: $$\frac{1}{2}\int_1^\infty u^{-1} du = \frac{1}{2}[ (-1+1)]=0$$ Since the integral of the comparison function converges, so does the original function by the Comparison Test.

Conclusion

By applying the Comparison Test, since the integral of the comparison function \(\int_1^{\infty} \frac{1}{x^2} dx\) converges, the given integral: $$\int_{1}^{\infty} \frac{2+\cos x}{x^{2}} d x$$ also converges.

Key concepts

Convergence of Integrals

Understanding whether an integral converges or diverges is a fundamental part of analyzing improper integrals. An improper integral often has one or both limits approaching infinity or involves an integrand that becomes infinite within the integration limits.

For example, in the integral \( \int_{1}^{\infty} \frac{2+\cos x}{x^{2}} d x \), we are dealing with an infinite upper limit. To determine convergence, we check if the total area under the curve from 1 to infinity is finite.

If the area is finite, the integral is said to converge. If not, it diverges. The main idea is to see if the integral effectively sums to a finite number despite its infinite bounds. For practical use, knowing if an integral converges helps in understanding the behavior of the function it represents, especially in physics and engineering problems.

Comparison Test

The Comparison Test is a useful method for determining the convergence or divergence of an improper integral by comparing it to a simpler or more well-known integral.

• Choose a simpler function that closely resembles the original integrand.
• Ensure the simpler function bounds the original from above or below.

For instance, we compared \( \frac{2+\cos x}{x^2} \) with the known function \( \frac{1}{x^2} \). Since \( \cos x \) oscillates, its maximum value is 1, establishing that \( \cos x \geq -1 \). This means the function \( \frac{2+\cos x}{x^2} \geq \frac{1}{x^2} \).

The Comparison Test states that if \( \int_{1}^{\infty} \frac{1}{x^2} d x \) converges, then \( \int_{1}^{\infty} \frac{2+\cos x}{x^{2}} d x \) also converges, providing a decisive conclusion without directly solving the more complex integral.

Integration Techniques

Choosing the right integration technique is crucial for evaluating integrals, especially improper ones. Typically, substitution, integration by parts, and partial fractions are employed depending on the form of the function.

In this case, the integral \( \int_{1}^{\infty} \frac{1}{x^2} d x \) is solved using substitution. By setting \( u = x^2 \), substituting for \( dx \), and simplifying, we transformed a potentially complex integration problem into a more straightforward exercise.

This approach highlights the importance of substitution in tackling improper integrals. It simplifies the evaluation and often converts difficult limits into easy ones, allowing us to confirm the behavior of integrals as approaching infinite bounds.

Always carefully select the integration method that reduces complexity and suits the specific features of the function you are investigating.