Question

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

Short answer

Answer: The integral diverges.

Step-by-step solution

Consider a simple comparison function

Since we want a function that behaves similarly to our integrand for large x values and converges, we can consider \(\frac{1}{\sqrt{x}}\). This is because the cosine term will be bound between -1 and 1 and won't affect the convergence for large x values.

Check if our comparison function converges or diverges

We can determine if our comparison function converges or diverges by evaluating its integral on the given interval: $$\int_{1}^{\infty}\frac{1}{\sqrt{x}}dx$$ To evaluate this improper integral, we will substitute a limit: $$\lim_{t\to\infty}\int_{1}^{t}\frac{1}{\sqrt{x}}dx$$ Now, we can find an antiderivative: $$\lim_{t\to\infty}(2\sqrt{x}\Big|_{1}^{t}) = \lim_{t\to\infty}(2\sqrt{t} - 2)$$ Since the limit as t goes to infinity of this expression is infinity, our comparison function diverges.

Compare the original integrand to the comparison function

Since the cosine term is bound between -1 and 1, our original integrand is less or equal to our comparison function. We can express this as: $$0 \leq \frac{2+\cos x}{\sqrt{x}} \leq \frac{1+\cos x + 1}{\sqrt{x}} = \frac{2}{\sqrt{x}}$$

Apply the comparison test

Since our comparison function \(\frac{1}{\sqrt{x}}\) diverges and the original integrand \(\frac{2+\cos x}{\sqrt{x}}\) is less than or equal to twice the comparison function, it follows that the original integral must diverge as well. We can therefore conclude that: $$\int_{1}^{\infty} \frac{2+\cos x}{\sqrt{x}} d x$$ diverges.

Key concepts

Improper Integral

An improper integral is a type of definite integral where either the interval of integration is infinite or the function being integrated has one or more points of discontinuity on the interval. Improper integrals are key in understanding the behavior of functions over infinite intervals or at points of discontinuity.

For example, in the exercise provided, the integrand \(\frac{2+\cos x}{\sqrt{x}}\) is not problematic in terms of discontinuity, but the interval of integration from 1 to infinity makes it an improper integral. To evaluate improper integrals, we often turn to limit processes, where we replace the infinity with a variable, say \(t\), and evaluate the limit as \(t\) approaches infinity.

Comparison Test

The comparison test is a method used to determine the convergence or divergence of improper integrals. It involves comparing the integral in question to another integral whose behavior (convergence or divergence) is known.

In our exercise, a comparison is made between the given integral and \(\int_{1}^{\infty}\frac{1}{\sqrt{x}}dx\), a simpler integral. By establishing that \(\frac{2+\cos x}{\sqrt{x}}\) is always less or equal to \(\frac{2}{\sqrt{x}}\), and that the integral of \(\frac{1}{\sqrt{x}}\) over the given range diverges, we apply the comparison test. It allows us to conclude that if the known integral diverges, and the integral in question is larger, then the integral in question must also diverge. This is a crucial step in determining the behavior of the original integral.

Limits

In calculus, limits are used to describe the behavior of a function as it approaches a certain point, without necessarily reaching that point. Limits are fundamental in defining derivatives, antiderivatives, and improper integrals.

Limit processes enable us to evaluate the behavior of functions at infinity or at points of discontinuity. In our exercise, the limit \(\lim_{t\to\infty}(2\sqrt{t} - 2)\) is crucial for determining the divergence of the comparison integral. By applying the limit, we are able to navigate the issue of the infinite range and draw conclusions about the behavior of the function as \(t\) becomes indefinitely large.

Antiderivatives

Antiderivatives, also known as indefinite integrals, represent the inverse operation to differentiation. When we find an antiderivative of a function, we're looking for a function whose derivative is the original function. These are essential in the solution of definite integrals, particularly when they involve limit processes.

In the context of our exercise, finding the antiderivative, \(2\sqrt{x}\), allows us to evaluate the integral from 1 to \(t\), and then take the limit as \(t\) approaches infinity. It's critical to get comfortable with antiderivatives since they are frequently used in solving a wide array of problems in calculus, from areas under curves to the dynamics of physical systems.