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Converge. Evaluate the integrals without using tables. $$\int_{-\infty}^{-2} \frac{2 d x}{x^{2}-1}$$

Short Answer

Expert verified
The integral converges to \(-\ln 3\).

Step by step solution

01

Identify the integral

The integral given is \( \int_{-\infty}^{-2} \frac{2 dx}{x^2 - 1} \). We need to evaluate this improper integral by breaking it down.
02

Simplify the integrand

The integrand \( \frac{2}{x^2 - 1} \) can be simplified using partial fraction decomposition. We rewrite the expression as \( \frac{2}{(x-1)(x+1)} \). Set \( \frac{2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \) and solve for \( A \) and \( B \).
03

Find partial fractions

Multiplying through by \( (x-1)(x+1) \), we have: \( 2 = A(x+1) + B(x-1) \).Expanding this gives \( 2 = (A+B)x + (A-B) \). To find \( A \) and \( B \), set the coefficients of like terms equal, yielding:\( A + B = 0 \) and \( A - B = 2 \).Solving, we find \( A = 1 \), \( B = -1 \).
04

Rewrite the integral

Using the values of \( A \) and \( B \), rewrite the integral as:\[ \int \frac{1}{x-1} - \frac{1}{x+1} \, dx = \int \left( \frac{1}{x-1} - \frac{1}{x+1} \right) dx. \]
05

Integrate the partial fractions

Integrate each term separately:\( \int \frac{1}{x-1} \, dx = \ln |x-1| + C_1 \) and \( \int \frac{-1}{x+1} \, dx = -\ln |x+1| + C_2 \).Thus, the integral becomes \( \ln |x-1| - \ln |x+1| + C \).
06

Evaluate the improper integral

Now, evaluate the improper integral from \( -\infty \) to \( -2 \):First, we take the limit:\[\lim_{t \to -\infty} \left( \ln |t-1| - \ln |t+1| \right) - \left( \ln |-2-1| - \ln |-2+1| \right).\]Computing the limits, we find:\( \lim_{t \to -\infty} (\ln |t-1| - \ln |t+1|) = 0 \) (since both go to \(-\infty\)).Calculate for \( x = -2 \):\( \ln |-3| - \ln |-1| = \ln 3 - \ln 1 = \ln 3 \).The result is \(- \ln 3 \).
07

Interpret the result

The integral converges to \(- \ln 3 \), indicating a finite value.

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Key Concepts

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

Partial Fraction Decomposition
Partial fraction decomposition is a powerful technique utilized in calculus to simplify rational expressions, making them easier to integrate. The process involves expressing a complex fraction as a sum of simpler fractions, each of which can be integrated individually.

To illustrate this, consider the fraction \( \frac{2}{x^2 - 1} \). This can be decomposed into simpler fractions of the form \( \frac{A}{x-1} + \frac{B}{x+1} \). By resolving the identity, the unknown coefficients \( A \) and \( B \) are determined through algebraic manipulation.
  • Multiply through by the original denominator: \( 2 = A(x+1) + B(x-1) \).
  • Expand and equate coefficients from both sides of the equation to solve for \( A \) and \( B \).
  • Substituting \( A = 1 \) and \( B = -1 \) gives us the separation into partial fractions.
Once the original fraction is expressed as a sum of simpler fractions, it becomes significantly more manageable to integrate each term separately.
Convergence of Integrals
Understanding the convergence of integrals, especially improper integrals, is crucial in evaluating them correctly. An integral is considered 'improper' when it involves infinite limits or discontinuous integrands. For these integrals, the limits must be evaluated to establish whether the integral converges to a finite value, or diverges, meaning it does not have a finite result.

In our problem, we have an integral from \(-\infty\) to \(-2\), which fits the definition of an improper integral due to the infinite lower limit. To test for convergence, we substitute these limits with a variable and take the limit to assess the behavior of the integral:
  • Compute \( \lim_{t \to -\infty} \left( \ln |t-1| - \ln |t+1| \right) \).
  • If this limit results in a finite number, the integral converges.
Here, as the terms approach infinity, they negate each other, leading the integral to converge to a value of \(- \ln 3 \), which signifies a definitive and finite result.
Limits in Calculus
Limits are a fundamental concept in calculus that deal with the behavior of functions as they approach specific points or as their inputs approach infinity. This concept is especially important in understanding and evaluating improper integrals.

When considering limits, we typically aim to find the value that a function approaches as the input gets closer to a certain number. In the context of improper integrals, limits are used to evaluate integrals with infinite boundaries:
  • For instance, in \( \lim_{t \to -\infty} \left( \ln |t-1| - \ln |t+1| \right) \), seeing how each expression behaves when \( t \) approaches \(-\infty\) is crucial.
  • Understanding that \( \ln |t-1| \) and \( \ln |t+1| \) cancel out each other's tendencies, allows us to conclude the evaluation accurately.
Grasping limits in calculus empowers you to predict the behavior of functions and to confirm the convergence or divergence of improper integrals by showing the resultant finite or infinite value.

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