Problem 5 - Calculus - AP Edition

Question

Evaluate the integrals that converge. $$\int_{3}^{+\infty} \frac{2}{x^{2}-1} d x$$

Step-by-Step Solution

Answer

The integral diverges.

Step 1: Identify the Type of Integral

The given integral is improper because the upper limit is infinity. We need to determine whether this integral converges.

Step 2: Express as a Limit

To evaluate the improper integral, we express it as a limit:\[ \lim_{b \to +\infty} \int_{3}^{b} \frac{2}{x^{2}-1} \, dx \]

Step 3: Simplify the Integrand

The denominator can be factored using the difference of squares: \(x^2 - 1 = (x-1)(x+1)\). Let's use partial fractions to split the fraction into simpler terms. This requires finding \(\frac{2}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1}\).

Step 4: Find the Partial Fraction Decomposition

Multiply through by \((x-1)(x+1)\) to remove the fractions:\[2 = A(x+1) + B(x-1)\]. Expand and solve for \(A\) and \(B\). By substituting values, we find \(A = 1\) and \(B = 1\). Thus, \[ \frac{2}{x^2-1} = \frac{1}{x-1} + \frac{1}{x+1} \].

Step 5: Rewrite the Integral Using Partial Fractions

Substitute the partial fraction decomposition back into the integral:\[ \int_{3}^{b} \left( \frac{1}{x-1} + \frac{1}{x+1} \right) \, dx \]. This can be separated as two integrals:\[ \int_{3}^{b} \frac{1}{x-1} \, dx + \int_{3}^{b} \frac{1}{x+1} \, dx \].

Step 6: Integrate Each Term

Integrate each term separately:\[ \int \frac{1}{x-1} \, dx = \ln|x-1| + C_1 \] and \[ \int \frac{1}{x+1} \, dx = \ln|x+1| + C_2 \]. Evaluating from 3 to \(b\), we get:\[ \left[ \ln|x-1| \right]_{3}^{b} + \left[ \ln|x+1| \right]_{3}^{b} \].

Step 7: Apply the Fundamental Theorem of Calculus

Substitute the limits into the integrated expressions:\[ \ln|b-1| - \ln|3-1| + \ln|b+1| - \ln|3+1| \]. Simplify to:\[ \ln|b-1| + \ln|b+1| - \ln|2| - \ln|4| \].

Step 8: Evaluate the Limit

Since both \(|b-1|\) and \(|b+1|\) go to infinity as \(b\) goes to infinity, the expression \(\ln|b-1| + \ln|b+1|\) goes to infinity. Therefore, the limit diverges.

Key Concepts

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to express a complex fraction as a sum of simpler fractions. This method is especially useful when integrating rational functions. In the exercise, we have the function \( \frac{2}{x^2 - 1} \). The denominator is a product of two terms: \((x-1)(x+1)\). By using partial fraction decomposition, we rewrite the expression:

• Set \( \frac{2}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1} \)
• Clear out the denominators by multiplying through by \((x-1)(x+1)\)
• This leads to the equation \(2 = A(x+1) + B(x-1)\)

Next, solve for \(A\) and \(B\) by plugging in convenient values of \(x\) or by equating coefficients. Here, we find \(A = 1\) and \(B = 1\), resulting in:

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

This helps break down the original integral into two simpler integrals that are easier to solve.

Convergence of Integrals

The Convergence of Integrals refers to whether an improper integral results in a finite value. Determining this is crucial when dealing with limits of integration that extend to infinity, like in the given problem:

• The improper integral is \( \int_{3}^{+\infty} \frac{2}{x^2-1} \, dx \)

First, express the integral as a limit to evaluate convergence:

• \( \lim_{b \to +\infty} \int_{3}^{b} \left( \frac{1}{x-1} + \frac{1}{x+1} \right) \, dx \)

To analyze convergence, note that both terms \( \frac{1}{x-1} \) and \( \frac{1}{x+1} \) are undefined at their respective asymptotes. However, since we're integrating from 3 to \( b \), this isn't an immediate concern. What matters is the behavior as \( b \to +\infty \).Unfortunately, the result is that both expressions \( \ln|b-1| \) and \( \ln|b+1| \) grow without bound, leading to a divergent integral. This means that the original integral does not converge to a finite value.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, essentially asserting that they are inverse processes. It provides a method for evaluating definite integrals:

• If \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then:

\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]In this exercise, once we have decomposed the fraction and computed the antiderivatives, we apply the theorem directly:

• For \( \int \frac{1}{x-1} \, dx \), the antiderivative is \( \ln|x-1| \)
• For \( \int \frac{1}{x+1} \, dx \), it is \( \ln|x+1| \)

Using the theorem, evaluate these integrals from the bounds 3 to \( b \), yielding:

• \( \left[ \ln|x-1| \right]_{3}^{b} + \left[ \ln|x+1| \right]_{3}^{b} \)
• This evaluates as \( \ln|b-1| + \ln|b+1| - \ln|2| - \ln|4| \)

While this setup prepares the integral for analysis, in reality, both terms involving \( b \) approach infinity, resulting in divergence.