Problem 36 - Calculus Single Variable

Question

Determine whether the given improper integral converges or diverges. If it converges, then evaluate it. $$ \int_{-\infty}^{0} \frac{\ln (2-x)}{2-x} d x $$

Step-by-Step Solution

Answer

The integral diverges; it does not converge.

Step 1: Analyze the Integral

The given integral is \[ \int_{-\infty}^{0} \frac{\ln (2-x)}{2-x} \, d x \] which is an improper integral, as the lower limit of integration is \( -\infty \). We need to check for convergence by either evaluating the integral or proving its convergence.

Step 2: Substitute Variables

Introduce the substitution \( u = 2-x \), implying \( du = -dx \). The limits of integration change correspondingly; as \( x \to -\infty \), \( u \to \infty \), and as \( x = 0 \), \( u = 2 \). The transformed integral is \[ \int_{\infty}^{2} \frac{\ln u}{u} (-du) = \int_{2}^{\infty} \frac{\ln u}{u} \, du \] since the limits change sign and order.

Step 3: Determine Convergence

The integral becomes \[ \int_{2}^{\infty} \frac{\ln u}{u} \, du \] which can be examined for convergence. A common technique is to compare it to another known integral, such as \( \int_{2}^{\infty} \frac{1}{u} \, du \), which diverges. Since \( \ln u > 1 \) for \( u > 2 \), and because \( \frac{\ln u}{u} > \frac{1}{u} \), it follows that our integral diverges.

Step 4: Conclusion

Since the integral \( \int_{2}^{\infty} \frac{\ln u}{u} \, du \) diverges by comparison with the harmonic series integral \( \int_{2}^{\infty} \frac{1}{u} \, du \), the original integral diverges as well.

Key Concepts

Convergence and Divergence

When dealing with improper integrals, it's important to determine whether they converge or diverge. Understanding these terms is key:

• Converge: An integral converges if it approaches a finite number as the limit of integration extends to infinity or a discontinuous point.
• Diverge: An integral diverges if it grows without bound or does not approach a finite limit.

To analyze convergence, we often compare the given integral to a simpler one with known behavior, like the harmonic series \( \int_{a}^{\infty} \frac{1}{x} \, dx \). If our integral's behavior is similar, we can draw conclusions about its convergence or divergence. In our example, you notice that the integral involving \( \frac{\ln u}{u} \) diverges by comparison because the logarithmic function increases more slowly than a linear function, but it still contributes to the divergence over infinite bounds.

Integration Techniques

Improper integrals often require specialized techniques to assess their nature. For this specific case, the techniques center around:

• Comparison Test: This method involves comparing the original integral to another integral with known convergence/divergence characteristics.
• Change of Limits: Adjusting the limits helps simplify the evaluation, especially when working with infinite bounds.

The comparison test especially helps us infer the behavior of our targeted integral. Our original function \( \frac{\ln(2-x)}{2-x} \) was transformed into \( \frac{\ln u}{u} \). By comparing this transformed function with \( \frac{1}{u} \), a function known to diverge on \( [2, \infty) \), we can conclude that the original integral also diverges. This is because the \( \ln u \) component does not diminish fast enough at infinity to counteract the divergence, showing that advanced integration techniques are crucial for understanding the behavior of improper integrals.

Substitution Method

The substitution method is a powerful tool to simplify integrals. By converting the given integral into a friendlier form, it becomes easier to evaluate or infer properties. In our problem, we used:

• Substituting Variables: Choose a substitution that simplifies the integral's expression. Here, setting \( u = 2-x \) helps transform the integral into terms easier to analyze.
• Reversing Limits: When substituting, the limits change according to the substitution's definitions. Thus, the limits reverse as \( x \to -\infty \to u \to \infty \) and \( x = 0 \to u = 2 \).

Incorporating substitution helps restructure the integral's domain and range, providing clarity on its convergence. After substitution, our original integral morphs into \( \int_{2}^{\infty} \frac{\ln u}{u} \, du \), which makes it apparent that evaluating the behavior of \( \ln u \) over \( u \) takes the forefront, ultimately leading to insights into its divergence. This method underscores the power of substitution in tackling improper integrals.