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Calculate the integrals in Exercises \(5-33,\) if they converge. You may calculate the limits by appealing to the dominance of one function over another, or by l'Hopital's rule. $$\int_{4}^{\infty} \frac{d x}{x^{2}-1}$$

Short Answer

Expert verified
The integral converges with a value of \(-\frac{1}{2} \ln\left(\frac{3}{5}\right)\).

Step by step solution

01

Analyze the Function for Convergence

The integral is given as \( \int_{4}^{\infty} \frac{dx}{x^2 - 1} \). Before attempting to evaluate it, we need to determine if it converges or diverges. The behavior of \( \frac{1}{x^2 - 1} \) for large \( x \) should be compared to known functions that converge or diverge. For large values of \( x \), the function \( \frac{1}{x^2 - 1} \) behaves similarly to \( \frac{1}{x^2} \), which is a known convergent function for \( x > 1 \). Therefore, the integral is likely convergent.
02

Set Up the Integral for Evaluation

Now, set up the limit for the improper integral: \[ \lim_{b \to \infty} \int_{4}^{b} \frac{dx}{x^2 - 1}. \] This expression will allow us to evaluate the integral from 4 to infinity.
03

Use Partial Fractions

Since \( x^2 - 1 \) can be factored as \( (x-1)(x+1) \), we will use partial fractions to decompose \( \frac{1}{x^2 - 1} \). Assume \( \frac{1}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1} \). Solving for \( A \) and \( B \) yields \( A = \frac{1}{2} \) and \( B = -\frac{1}{2} \). Thus, \[ \frac{1}{x^2 - 1} = \frac{1}{2(x-1)} - \frac{1}{2(x+1)}. \]
04

Integrate and Evaluate the Limit

Insert the decomposed form into the integral: \[ \lim_{b \to \infty} \int_{4}^{b} \left( \frac{1}{2(x-1)} - \frac{1}{2(x+1)} \right) dx. \] Integrate term by term to get: \[ \lim_{b \to \infty} \left[ \frac{1}{2} \ln|x-1| - \frac{1}{2} \ln|x+1| \right]_{4}^{b}. \] Simplifying gives \[ \lim_{b \to \infty} \frac{1}{2} \left( \ln\frac{|b-1|}{|b+1|} \right) - \frac{1}{2} \left( \ln\frac{3}{5} \right). \] As \( b \to \infty \), \( \ln\frac{b-1}{b+1} \to 0 \). Thus, the integral evaluates to \(-\frac{1}{2} \ln\left(\frac{3}{5}\right)\).
05

Conclusion on Convergence and Result

Since we were able to evaluate the integral and find a finite result, the integral converges. The final value of the integral is \(-\frac{1}{2} \ln\left(\frac{3}{5}\right)\).

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Key Concepts

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

Convergence of Integrals
When tackling improper integrals, the first task is to determine if the integral converges or diverges. This involves evaluating the behavior of the function as it approaches infinity or a vertical asymptote. In our exercise, the expression given is \( \int_{4}^{\infty} \frac{dx}{x^2 - 1} \). To check for convergence, compare the function \( \frac{1}{x^2 - 1} \) to known convergent/divergent functions. For large values of \( x \), \( \frac{1}{x^2 - 1} \) closely resembles \( \frac{1}{x^2} \), a convergent function when \( x > 1 \). This initial analysis suggests that our integral is likely to converge.
It is essential to set up a limit as one of the bounds approaches infinity to properly evaluate an improper integral. The use of limits will allow us to manage the infinite bound effectively and confirm the convergence or divergence finding through further evaluation.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to simplify rational expressions, making them easier to integrate. This method breaks a complex fraction into simpler fractions. In this exercise, the given function is \( \frac{1}{x^2 - 1} \), which can be factored into \( (x-1)(x+1) \). The aim is to express \( \frac{1}{x^2 - 1} \) as \( \frac{A}{x-1} + \frac{B}{x+1} \).
This process involves solving for the constants \( A \) and \( B \), resulting in \( A = \frac{1}{2} \) and \( B = -\frac{1}{2} \). Thus, \( \frac{1}{x^2 - 1} = \frac{1}{2(x-1)} - \frac{1}{2(x+1)} \). This decomposition facilitates carrying out the integration process because it reduces the function into simpler terms that integrate directly using basic logarithmic rules.
L'Hopital's Rule
L'Hopital's Rule is a useful tool for evaluating limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In the context of improper integrals, it can occasionally be helpful when assessing convergence or calculating the limits of the resulting expressions.
Although L'Hopital's Rule was not directly employed in this exercise, it is a valuable concept to understand. Being able to recognize conditions under which L'Hopital's Rule can be applied—with the differentiation of the numerator and denominator—is important for problems involving limits and can simplify complex limit evaluations.
Limits of Functions
In calculus, the concept of limits is fundamental for understanding the behavior of functions as they approach certain points. In improper integrals, limits are crucial for analyzing what happens as the variable approaches infinity.
  • In our exercise, we set up the limit for \( \int_{4}^{\infty} \frac{dx}{x^2 - 1} \) as \( b \to \infty \).
  • Limits allow us to handle infinite bounds effectively. Calculating these helps us find the result of the integral even when one of its boundaries is unbounded.
This exercise concludes that the limit also demonstrates convergence as \( b \to \infty \), where terms simplify to \( \ln \frac{b-1}{b+1} \to 0 \). Therefore, the convergence and the finite result \(-\frac{1}{2} \ln\left(\frac{3}{5}\right)\) confirm our integral converges, and the limit is essential for understanding its behavior at infinity.

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