Problem 72 - Calculus Single Variable

Question

Use the Comparison Theorem to establish that the given improper integral is divergent. $$ \int_{1}^{\infty} \frac{3-\sin (x)}{2 x-1} d x $$

Step-by-Step Solution

Answer

The integral diverges by comparison to \( \int_{1}^{\infty} \frac{2}{2x-1} \, dx \).

Step 1: Identify the Function and Limits

We are tasked with evaluating the improper integral \( \int_{1}^{\infty} \frac{3-\sin(x)}{2x-1} \, dx \). This integral has an infinite upper limit, which is where the potential divergence comes into play.

Step 2: Determine the Dominant Function

To use the Comparison Theorem, we need to compare \( \frac{3-\sin(x)}{2x-1} \) to a simpler function. Notice that \(-1 \leq \sin(x) \leq 1\), which implies \(2 \leq 3 - \sin(x) \leq 4\). Thus, \( \frac{2}{2x-1} \leq \frac{3-\sin(x)}{2x-1} \leq \frac{4}{2x-1} \). We will compare our function to \( \frac{2}{2x-1} \).

Step 3: Evaluate the Comparison Function

Consider the integral \( \int_{1}^{\infty} \frac{2}{2x-1} \, dx \). Perform the substitution \( u = 2x - 1 \), which gives \( du = 2 \, dx \), or \( dx = \frac{du}{2} \). The limits of integration change accordingly from \( x=1 \) to \( u=1 \) and from \( x=\infty \) to \( u=\infty \). This transforms the integral into \( \frac{1}{2} \int_{1}^{\infty} \frac{2}{u} \, du = \int_{1}^{\infty} \frac{1}{u} \, du \).

Step 4: Test for Divergence

The integral \( \int_{1}^{\infty} \frac{1}{u} \, du \) is a known divergent integral (a p-integral with \( p=1 \)), which implies that \( \int_{1}^{\infty} \frac{2}{2x-1} \, dx \) also diverges. By the Comparison Theorem, since \( \frac{3-\sin(x)}{2x-1} \geq \frac{2}{2x-1} \) and the integrated comparison diverges, the original integral \( \int_{1}^{\infty} \frac{3-\sin(x)}{2x-1} \, dx \) is also divergent.

Key Concepts

Comparison Theorem

The Comparison Theorem is a valuable tool used to determine the convergence or divergence of improper integrals. It works by comparing a complex integral to a simpler one, for which the behavior is well known. The idea is, if you have two functions, say \( f(x) \) and \( g(x) \), where \( 0 \leq f(x) \leq g(x) \) for all \( x \) in the domain of interest, and if \( \int g(x) \, dx \) diverges, then \( \int f(x) \, dx \) also diverges.

In the given exercise, you must analyze the integral \( \int_{1}^{\infty} \frac{3-\sin(x)}{2x-1} \, dx \). The function \( \frac{3-\sin(x)}{2x-1} \) is more complex. So, we compare it to another function, \( \frac{2}{2x-1} \), which is simpler and dominates the behavior of the original function over the interval of integration. By establishing the inequality \( \frac{2}{2x-1} \leq \frac{3-\sin(x)}{2x-1} \leq \frac{4}{2x-1} \), we can apply the Comparison Theorem. If the simpler integral diverges, our original integral also diverges, confirming the solution using this theorem.

Divergence

Divergence in the context of integrals refers to a situation where an integral does not settle to a finite value. This occurs in improper integrals when the limit does not exist or is infinite. In the exercise, you've seen the integral \( \int_{1}^{\infty} \frac{1}{u} \, du \). This is a classic example of a divergent integral because it does not converge to a finite number as \( u \) approaches infinity.

When dealing with improper integrals, recognizing divergence is crucial because it can immediately conclude that further evaluation will not yield a finite result. Divergence is identified if, as you approach the bounds of integration, the function does not taper off but instead heads toward infinity or oscillates without settling. This property is a cornerstone in proving divergence using the Comparison Theorem: show that the comparison function diverges, and thus, so does the original integral.

p-Integrals

p-Integrals are a specific kind of integral of the form \( \int_{a}^{b} \frac{1}{x^p} \, dx \). Whether they converge or diverge depends on the value of \( p \) and the limits of integration. A commonly referenced case is \( \int_{1}^{\infty} \frac{1}{x} \, dx \) where \( p = 1 \), which is known to diverge.

In the exercise, after transforming and simplifying the integral using substitution, it becomes clear how a p-integral plays a role. The integral \( \int_{1}^{\infty} \frac{1}{u} \, du \) shows how \( p = 1 \) causes this specific p-integral to diverge. In general:

• If \( p > 1 \), the improper integral \( \int_{1}^{\infty} \frac{1}{x^p} \, dx \) converges.
• If \( p \leq 1 \), it diverges.

Understanding p-integrals helps in properly evaluating and predicting the behavior of more complex integrals, like those seen with the Comparison Theorem and in divergence assessments. Thus, they serve as a guideline for other similar integrals.