Problem 29 - Applied Calculus

Question

17-40. Evaluate each improper integral or state that it is divergent. $$ \int_{0}^{\infty} \frac{x}{x^{2}+1} d x $$

Step-by-Step Solution

Answer

The integral is divergent.

Step 1: Identify the Type of Improper Integral

We are given \(\int_{0}^{\infty} \frac{x}{x^{2}+1} \ dx\). This is an improper integral because the upper limit of integration is infinity. To tackle this, we need to take the limit as \( b \to \infty \).

Step 2: Set Up the Limit Definition of the Improper Integral

Rewrite the integral with a finite upper limit, and take the limit as this upper limit approaches infinity: \[ \int_{0}^{\infty} \frac{x}{x^{2}+1} \ dx = \lim_{b \to \infty} \int_{0}^{b} \frac{x}{x^{2}+1} \ dx. \]

Step 3: Use Substitution Technique

Let us use the substitution \( u = x^2 + 1 \), which implies \( du = 2x \, dx \). Therefore, \( x \, dx = \frac{1}{2} \, du \). The limits change accordingly: when \( x = 0, \, u = 1 \) and when \( x = b, \, u = b^2 + 1 \).

Step 4: Rewrite the Integral in Terms of the Substitution

The integral becomes: \[ \frac{1}{2} \int_{1}^{b^2+1} \frac{1}{u} \, du. \] This can be integrated to yield: \[ \frac{1}{2} \left[ \ln |u| \right]_{1}^{b^2+1}. \]

Step 5: Evaluate the Integral

Evaluate the antiderivative at the limits: \[ \frac{1}{2} \left[ \ln(b^2+1) - \ln(1) \right] = \frac{1}{2} \ln(b^2+1). \]

Step 6: Take the Limit as b Approaches Infinity

Now take the limit: \[ \lim_{b \to \infty} \frac{1}{2} \ln(b^2+1). \] As \( b \to \infty \), \( \ln(b^2+1) \to \infty \), hence the integral diverges.

Key Concepts

Substitution Technique

In the realm of calculus, improper integrals often appear intimidating. A powerful tool to simplify them is the substitution technique. This mathematical "trick" helps us transform a complicated integral into an easier one. Won't that be useful? In our given problem, we have the improper integral \( \int_{0}^{\infty} \frac{x}{x^{2}+1} \ dx \), featuring an infinite upper limit. To handle this, we apply a substitution.Let's look at how substitution plays its role:

• We choose \( u = x^2 + 1 \). This choice simplifies our integrand because \( u \) is the expression in the denominator.
• Calculating the derivative, we find \( du = 2x \, dx \) or rearranged, \( x \, dx = \frac{1}{2} \, du \).
• Substituting these into our integral, the expression transforms significantly. Although the original integration seems tough, the substitution simplifies it to \( \frac{1}{2} \int_{1}^{b^2+1} \frac{1}{u} \, du \).

This substitution technique converts a difficult integral into a simpler one, where solving becomes more straightforward. Remember, choosing the right substitution is like finding the right key to a lock.

Limits of Integration

Dealing with improper integrals, especially when infinity comes into play, requires deep understanding of limits of integration. It appears prominently in our exercise, where we were asked to compute the integral from 0 to infinity of \( \frac{x}{x^{2}+1} \).Here's how we handled it:

• We started by recognizing that \( \infty \) represented as a limit rather than a number. We expressed the integral as \( \lim_{b \to \infty} \int_{0}^{b} \frac{x}{x^{2}+1} \ dx \).
• This process allows us to focus only on integrals with finite boundaries. It creates a finite problem that gradually approaches the original infinite one as \( b \) grows.

By understanding limits of integration, we systematically approach infinite boundaries. The value of the integral depends greatly on this concept, making it a cornerstone in evaluating improper integrals efficiently.

Divergence of Integrals

The notion of divergence tells us that an integral may not always yield a finite number. Specifically, in the given exercise, we explored the divergence of the integral \( \int_{0}^{\infty} \frac{x}{x^{2}+1} \ dx \).Here's how divergence was identified:

• After integrating using substitution, we evaluated the expression \( \frac{1}{2} \ln(b^2+1) \).
• As \( b \to \infty \), the logarithm term \( \ln(b^2+1) \) approached infinity, leading our integral to diverge.
• This result indicates that the area under the curve is infinite, which is a classic sign of divergence.

Recognizing divergence early is crucial as it suggests that solutions don't exist in the realms of boundedness or finiteness. Understanding when, why, and how an integral diverges forms a key aspect of mastering improper integrals.