Problem 12 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The integral \( \int_{2}^{\infty} \frac{2 \, d t}{t^{2}-1} \) is divergent.

Step 1: Identify the Integral Type

The integral \( \int_{2}^{\infty} \frac{2 \, d t}{t^{2}-1} \) is an improper integral because the upper limit is infinity. We need to evaluate it to determine if it is convergent or divergent.

Step 2: Simplify the Integrand

Simplify the integrand \( \frac{2}{t^2 - 1} \) by recognizing that it can be decomposed using partial fraction decomposition. Notice that \( t^2 - 1 = (t - 1)(t + 1) \), so we can express \( \frac{2}{t^2 - 1} \) as \( \frac{A}{t-1} + \frac{B}{t+1} \).

Step 3: Determine Coefficients for Partial Fractions

Set \( \frac{2}{t^2 - 1} = \frac{A}{t-1} + \frac{B}{t+1} \). Multiply through by \( t^2 - 1 \) to get \( 2 = A(t+1) + B(t-1) \). Solve for \( A \) and \( B \) by setting up equations: \( A + B = 0 \) and \( A - B = 2 \). Solving these gives \( A = 1 \) and \( B = -1 \).

Step 4: Write the Integral with Partial Fractions

Rewrite the integral using the partial fractions: \( \int_{2}^{\infty} \frac{1}{t-1} \, d t - \int_{2}^{\infty} \frac{1}{t+1} \, d t \).

Step 5: Evaluate the Integral as Two Separate Improper Integrals

Evaluate each integral separately. The first integral is \( \int_{2}^{\infty} \frac{1}{t-1} \, d t \) which is an improper integral requiring a limit. Similarly, evaluate \( \int_{2}^{\infty} \frac{1}{t+1} \, d t \).

Step 6: Solve the Improper Integrals

For \( \int_{2}^{\infty} \frac{1}{t-1} \, d t \), use the limit: \( \lim_{b \to \infty} \left[ \ln |t - 1| \right]_2^b \). For the second integral, \( \lim_{b \to \infty} \left[ \ln |t + 1| \right]_2^b \). Compute both limits, both diverge because they approach infinity.

Step 7: Conclusion of Divergence

Since both integrals from Step 6 diverge (approach infinity), the original integral \( \int_{2}^{\infty} \frac{2 \, d t}{t^2 - 1} \) is also divergent.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler ones, making them easier to integrate. When faced with a rational integrand, like in this exercise, partial fractions allow us to express it as a sum of simpler fractions.

• Start by factoring the denominator. In our case, the denominator is \( t^2 - 1 \), which can be factored into \((t - 1)(t + 1)\).
• Express the original fraction \(\frac{2}{t^2 - 1}\) as the sum of two separate fractions: \(\frac{A}{t - 1} + \frac{B}{t + 1}\).
• To find \(A\) and \(B\), multiply both sides by the original denominator and equate coefficients, resulting in a system of equations. Here, \(A = 1\) and \(B = -1\).

This decomposition simplifies solving the integral, transforming it into two simpler integrals that can be tackled individually.

Convergent and Divergent Integrals

Determining whether an integral is convergent or divergent is crucial in evaluating improper integrals. Convergent integrals have a finite value, while divergent integrals do not have a limit and can approach infinity.
To identify if an integral converges or diverges:

• Examine the behavior of the integrand as it approaches the boundaries of integration. If both individual integrals diverge when evaluated from the lower limit to infinity, the original integral will also diverge.
• Comparing the integrand to known integrals or using the comparison test can provide insight into whether the integral converges or diverges.
• In our case, after rewriting the integral using partial fractions, both integrals \(\int_{2}^{\infty} \frac{1}{t-1} \, d t\) and \(\int_{2}^{\infty} \frac{1}{t+1} \, d t\) diverge. Therefore, the integral of interest is divergent as well.

Understanding convergence and divergence helps predict the behavior of integrals over infinite intervals or unbounded functions.

Limit of Integration

The limit of integration characterizes the endpoints of the interval over which we are integrating. In improper integrals, if one or both of these limits extend to infinity, a special approach is needed to compute the integral.
For an integral such as \(\int_{2}^{\infty} f(t) \, dt\), we replace the upper limit of integration with a variable, typically \(b\), that tends toward infinity:

• Rewrite the integral as \(\int_{2}^{b} f(t) \, dt\).
• Calculate this integral symbolically as if \(b\) were finite.
• Evaluate the limit as \(b\) approaches infinity to assess convergence.

In step 6 of our exercise, we derived the limits for both \(\int_{2}^{\infty} \frac{1}{t-1} \, dt\) and \(\int_{2}^{\infty} \frac{1}{t+1} \, dt\), observing their divergence since both logarithmic expressions approached infinity.

Infinite Integration

When integrating over a range that extends to infinity, we perform what is termed infinite integration. The focus is not only on evaluating the integral but also on checking its behavior at the upper boundary of integration.
Steps to handle infinite integration include:

• Replace infinite limits with a variable and approach these limits using the limit process.
• For integrals such as \(\int_{a}^{\infty} f(t) \, dt\), we calculate \(\lim_{b \to \infty} \int_{a}^{b} f(t) \, dt\).
• Analyzing the convergence or divergence of these calculated limits refines the solution.

In our example, we treated both parts of the decomposed integral this way, recognizing the divergence of each as their respective logarithmic evaluations extended to infinity, thus leading to the conclusion that the original integral was divergent.