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Chapter 7: Problem 4

Evaluate the integrals that converge. $$ \int_{-1}^{+\infty} \frac{x}{1+x^{2}} d x $$

Short Answer

Expert verified
The integral diverges.

Step by step solution

01

Identify the Type of Integral

The given integral is \( \int_{-1}^{+\infty} \frac{x}{1+x^2} \, dx \). Because it has an infinite upper limit, it's considered an improper integral.
02

Set Up the Improper Integral

We rewrite the integral with a limit to properly handle the infinite boundary: \[ \lim_{b \to +\infty} \int_{-1}^{b} \frac{x}{1+x^2} \, dx \]
03

Evaluate the Integral

To evaluate \( \int \frac{x}{1+x^2} \, dx \), use a substitution. Let \( u = 1 + x^2 \), then \( du = 2x dx \) or \( x dx = \frac{1}{2} du \). The integral becomes:\[ \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C \]Substitute back for \( u \):\[ \frac{1}{2} \ln |1+x^2| + C \]
04

Evaluate the Definite Integral

Using the result of the indefinite integral, calculate:\[ \lim_{b \to +\infty} \left[ \frac{1}{2} \ln |1+x^2| \right]_{-1}^{b} = \lim_{b \to +\infty} \left( \frac{1}{2} \ln(1+b^2) - \frac{1}{2} \ln(2) \right) \]Since the term \( \ln(1+b^2) \) tends towards infinity as \( b \to +\infty \), the integral diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Process
When evaluating improper integrals, the limit process is a central technique. Improper integrals usually involve infinite limits or unbounded integrands, making them impossible to evaluate directly.
Instead, we replace the problematic boundaries with a limit. Take, for example, the integral\( \int_{-1}^{+\infty} \frac{x}{1+x^2} \, dx \). Here, the upper limit is infinite, which transforms the integral into an improper one.
To properly handle this, we introduce a variable, like \( b \), to replace infinity.
  • Set up: \( \lim_{b \to +\infty} \int_{-1}^{b} \frac{x}{1+x^2} \, dx \)
  • The limit \( b \to +\infty \) allows us to consider the behavior of the integral as it extends towards infinity.
This limit process is fundamental for determining if an improper integral converges (approaches a finite value) or diverges (grows indefinitely).
Substitution Method
The substitution method is a powerful tool for simplifying and evaluating integrals. It involves changing variables to make an integral easier to solve.
In our exercise, we encounter the integral \( \int \frac{x}{1+x^2} \, dx \). At first glance, it seems complex, but substitution can simplify it.
Here's how you can use substitution:
  • Choose a substitution: Let \( u = 1 + x^2 \)
  • Find the differential of \( u \), which is \( du = 2x \, dx \)
  • Rewriting the integral with \( x \, dx = \frac{1}{2} \, du \) results in \( \frac{1}{2} \int \frac{1}{u} \, du \)
  • This simplifies to \( \frac{1}{2} \ln|u| + C \)
Substituting back for \( u \) gives \( \frac{1}{2} \ln|1+x^2| + C \). This method turns complicated expressions into more manageable forms, making the integral solvable.
Divergence of Integrals
Determining whether an integral converges or diverges is crucial in calculus, especially for improper integrals. Divergence occurs when the integral does not settle on a finite value.
In our problem, after applying the substitution method and limit process, we find:
\( \lim_{b \to +\infty} \left( \frac{1}{2} \ln(1+b^2) - \frac{1}{2} \ln(2) \right) \)
The term \( \ln(1+b^2) \) grows unbounded as \( b \) approaches infinity.
  • This indicates that the integral's value increases indefinitely.
  • Ultimately, we conclude that the integral diverges.
Divergence signifies that no finite area exists under the curve for the given interval, highlighting the integral's infinite nature.

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