Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each improper integral or show that it diverges. $$ \int_{1}^{10} \frac{d x}{x \ln ^{100} x} $$

Step-by-Step Solution

Answer

The integral diverges.

Step 1: Verify the Nature of the Integral

First, we need to determine if the integral is improper. The integral \( \int_{1}^{10} \frac{dx}{x \ln^{100} x} \) is not improper at \( x = 1 \) or \( x = 10 \). Thus, it is not improper in terms of boundary limits.

Step 2: Analyze the Integrand

We look at the form of the integrand: \( \frac{1}{x \ln^{100} x} \). Since the integrand involves \( \ln x \) raised to a power, direct integration could be challenging. This suggests a substitution might simplify the evaluation.

Step 3: Use Substitution

Let \( u = \ln x \) which implies \( du = \frac{1}{x} dx \). The bounds of \( u \) when \( x = 1 \) is \( \ln 1 = 0 \) and when \( x = 10 \) is \( \ln 10 \). The integral becomes \( \int_{0}^{\ln 10} \frac{du}{u^{100}} \).

Step 4: Integrate Using Power Rule

Now, integrate the new expression using the power rule for integrals: \( \int u^n du = \frac{u^{n+1}}{n+1} + C \). For \( u^{-100} \), we get: \( \int u^{-100} du = \frac{u^{-99}}{-99} \).

Step 5: Evaluate the Definite Integral

Plug in the bounds of the integral: \( \left[ \frac{u^{-99}}{-99} \right]_{0}^{\ln 10} = \frac{(\ln 10)^{-99}}{-99} - \lim_{u \to 0} \frac{u^{-99}}{-99} \). Since \( u^{-99} \) goes to infinity as \( u \to 0 \), the limit diverges.

Step 6: Conclusion on Convergence

Since the limit when \( u \rightarrow 0 \) results in divergence, \( \int_{1}^{10} \frac{1}{x \ln^{100} x} dx \) diverges.

Key Concepts

Improper Integrals

In calculus, an integral is termed "improper" when it involves infinite limits or an integrand that becomes infinite within the limits of integration. Often, improper integrals challenge students as they require special techniques to evaluate them. For our exercise, we are given the integral \( \int_{1}^{10} \frac{dx}{x \ln^{100} x} \).
Though it appears that this integral might be improper due to its complex form, examining it shows it has finite boundaries, and the integrand does not approach infinity at the boundaries of 1 and 10. Hence, it is not classified as improper in terms of boundaries. However, boundaries aren't the only thing to consider; sometimes the behavior of the function itself within the integral needs more scrutiny to see if it forms an improper integral.

Integration Techniques

The process of integrating often involves more than direct application of formulas; it requires choosing suitable techniques to simplify problems. Here, identifying the core structure of our function\( \frac{1}{x \ln^{100} x} \) suggests substitution can help.
A common technique employed in integrating complex functions involves substitution. In our step-by-step solution:

• We opted for substituting \( u = \ln x \) based on the repetitive appearance of \( \ln x \).
• This substitution simplifies the complex integrand to a power function \( \frac{1}{u^{100}} \), which is easier to handle.
• The bounds naturally change with substitution, from \( x = 1 \) becoming \( u = 0 \), and \( x = 10 \) becoming \( u = \ln 10 \).

These changes drastically simplify the solving process by using the power rule integration on a transformed polynomial integral.

Integral Divergence

Divergence is a critical concept in determining the outcome of evaluating integrals. When an integral fails to settle to a finite number, it diverges, suggesting the area under the curve is infinite over the limit interval.
For the modified integral \( \int_{0}^{\ln 10} \frac{du}{u^{100}} \), as we evaluate its definite limits, the expression becomes\( \left[ \frac{u^{-99}}{-99} \right]_{0}^{\ln 10} \).
While \( (\ln 10)^{-99} \) gives a finite value, the potential cause for divergence arises at the lower bound where \( u \to 0 \).

• As \( u^{-99} \) approaches infinity, the expression becomes undefined, indicating divergence.
• Additionally, the negative exponent \( -99 \) causes the integral to diverge, showing the function does not settle to a finite area.
  Divergence stresses the importance of understanding the behavior of the integrand over the entire range, not just at endpoints.