Problem 9 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{-\infty}^{-2} \frac{2 d x}{x^{2}-1} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln(3) \).

Step 1: Identify the Type of Integral

The given integral is improper because it has an infinite limit of integration (-infinity) and a rational function that may have a vertical asymptote within its domain. We need to separately consider the behavior of the integrand as it approaches this vertical asymptote.

Step 2: Decompose the Integrand

The integrand is \( \frac{2}{x^2 - 1} \). This can be expressed as a partial fraction decomposition: \( \frac{2}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1} \). Solving for A and B by equating and factoring, we find \( A = 1 \) and \( B = -1 \). Hence, \[ \frac{2}{x^2 - 1} = \frac{1}{x-1} - \frac{1}{x+1}. \]

Step 3: Set Up the Limit for the Improper Integral

Rewrite the integral from \(-\infty\) to \(-2\) in terms of a limit: \[ \lim_{t \to -\infty} \int_{t}^{-2} \left( \frac{1}{x-1} - \frac{1}{x+1} \right) \, dx. \]

Step 4: Integrate the Decomposed Functions

Compute the antiderivative of \( \frac{1}{x-1} \) and \( \frac{1}{x+1} \). The antiderivative of \( \frac{1}{x-1} \) is \( \ln|x-1| \) and that of \( \frac{1}{x+1} \) is \( \ln|x+1| \). So, \[ \int \left( \frac{1}{x-1} - \frac{1}{x+1} \right) \, dx = \ln|x-1| - \ln|x+1| + C. \]

Step 5: Evaluate the Limit

Substitute back into the improper integral: \[ \lim_{t \to -\infty} \left[ \ln|-2-1| - \ln|-2+1| - (\ln|t-1| - \ln|t+1|) \right]. \] Simplifying, this becomes: \[ \lim_{t \to -\infty} \left( \ln(3) - \ln(1) - (\ln|t-1| - \ln|t+1|) \right). \] As \( t \to -\infty \), both \( \ln|t-1| \) and \( \ln|t+1| \) approach infinity, and their difference approaches zero, so we have \( \ln(3) \) as the final value.

Key Concepts

Partial Fraction Decomposition

When faced with integrating a complex rational function, partial fraction decomposition is a powerful tool. It breaks down a complicated fraction into simpler, more manageable components. For example, the integrand \( \frac{2}{x^2 - 1} \) is decomposed into the sum \( \frac{1}{x-1} - \frac{1}{x+1} \). This transformation is valid because we look for constants \( A \) and \( B \) that satisfy the equation:

• \( \frac{2}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1} \)

By solving for \( A \) and \( B \), generally by equating numerators and factoring, we simplify the integration process. In this case, it turns out \( A = 1 \) and \( B = -1 \), allowing us to split the original integrand into two simpler fractions, each easily integrable. This decomposition is especially useful in calculus because it turns a daunting computation into more approachable parts, each tractable by basic antiderivative rules.

Limits of Integration

Improper integrals are unique as they involve infinite limits. These require special handling, often involving limits in calculus. When an integral extends to infinity, it suggests potential division by zero or undefined behaviors at an endpoint.

In our example, notice the integration from

• \(-\infty \) to \(-2\).

This makes the integral improper. To address this, it is rewritten using the limit concept:

• \( \lim_{t \to -\infty} \int_{t}^{-2} \left( \frac{1}{x-1} - \frac{1}{x+1} \right) \ dx \)

Here, we substitute \( t \) for infinity and solve as \( t \) tends to negative infinity. This strategic move defines the integral over a finite interval, rendering it solvable.

Antiderivatives

Finding the antiderivatives is another crucial step in solving integrals. It involves determining a function whose derivative is the integrand. For the integrand components \( \frac{1}{x-1} \) and \( \frac{1}{x+1} \), we determine the following antiderivatives:

• \(\frac{1}{x-1}\) yields \(\ln|x-1|\)
• \(\frac{1}{x+1}\) yields \(\ln|x+1|\)

Thus, integrating the decomposed fractions sums up to \( \ln|x-1| - \ln|x+1| + C \), where \( C \) is the constant of integration. Antiderivatives are conceptual reversals of differentiation, a cardinal principle in integral calculus. This skill allows us to navigate through calculus problems of any complexity, turning abstract mathematical ideas into more concrete solutions.

Asymptotic Behavior

Understanding asymptotic behavior is crucial when dealing with improper integrals. In such contexts, examining how functions behave as they approach infinity or certain critical points helps determine the convergence or divergence of an integral.

While calculating

• \( \lim_{t \to -\infty} (\ln|t-1| - \ln|t+1|) \)

we observe that both terms tend toward infinity. Notably, their difference nears zero as they cancel out more with increasing \( |t| \). Recognizing this behavior, we deduce stability in the expression's limit.

In our problem, while individual logs diverge, their difference attains a finite result, contributing stability to otherwise unpredictable values. Asymptotic analysis thus offers insight into integral behavior over vast intervals, essential for digital computation and theoretical exploration.