Chapter 8: Problem 12
Evaluate each improper integral or show that it diverges. \(\int_{e}^{\infty} \frac{\ln x}{x} d x\)
Short Answer
Expert verified
The integral diverges.
Step by step solution
01
Analyze the Improper Integral
The given integral is \( \int_{e}^{\infty} \frac{\ln x}{x} \, dx \). An improper integral must be analyzed because the upper limit of integration is infinite. So we will substitute the upper limit with a variable \( t \) and then determines the limit as \( t \to \infty \).
02
Set Up the Limit Expression
We express the improper integral as a limit: \( \int_{e}^{\infty} \frac{\ln x}{x} \, dx = \lim_{t \to \infty} \int_{e}^{t} \frac{\ln x}{x} \, dx \).
03
Compute the Integral
To evaluate \( \int_{e}^{t} \frac{\ln x}{x} \, dx \), we recognize that \( \frac{\ln x}{x} \) is the derivative of \( \ln^2 x / 2 \). Thus, \( \int \frac{\ln x}{x} \, dx = \frac{1}{2}(\ln x)^2 + C \).
04
Evaluate the Definite Integral
Evaluate the definite integral from \( e \) to \( t \): \[\left[ \frac{1}{2} (\ln x)^2 \right]_{e}^{t} = \frac{1}{2} (\ln t)^2 - \frac{1}{2} (\ln e)^2.\] The value of \( \ln e \) is \( 1 \), so this simplifies to: \[\frac{1}{2} (\ln t)^2 - \frac{1}{2}.\]
05
Take the Limit as t Approaches Infinity
Now take the limit as \( t \to \infty \): \[ \lim_{t \to \infty} \left( \frac{1}{2} (\ln t)^2 - \frac{1}{2} \right).\] As \( t \) becomes very large, \( (\ln t)^2 \) also grows indefinitely, meaning the expression \( \frac{1}{2} (\ln t)^2 \) diverges to infinity.
06
Conclusion About Convergence or Divergence
Since the limit diverges to infinity, the initial improper integral \( \int_{e}^{\infty} \frac{\ln x}{x} \, dx \) also diverges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration Limits
Improper integrals often challenge learners due to their undefined limits, as they extend to infinity. In the integral \( \int_{e}^{\infty} \frac{\ln x}{x} \, dx \), the upper limit is infinity, presenting a classic case of an improper integral. To tackle such integrals:
- Substitute the infinite limit with a variable, usually denoted by \( t \).
- Then express the original integral as the limit of a more standard definite integral as \( t \rightarrow \infty \).
Convergence and Divergence
The core goal when dealing with improper integrals is determining whether they converge to a specific value or diverge, spiraling towards infinity. In step 5 of the original solution, we note that as \( t \) becomes very large, the term \( (\ln t)^2 \) grows unbounded.
It falls on the behavior of the integrand and that of the resulting expression after taking the limit to conclude if it converges or diverges.
Here, we see the entire expression \( \lim_{t \to \infty} \left( \frac{1}{2} (\ln t)^2 - \frac{1}{2} \right) \) clearly diverges since \( (\ln t)^2 \to \infty \). This divergence tells us that the original integral does not settle to a specific value but rather extends indefinitely, highlighting its divergence.
It falls on the behavior of the integrand and that of the resulting expression after taking the limit to conclude if it converges or diverges.
Here, we see the entire expression \( \lim_{t \to \infty} \left( \frac{1}{2} (\ln t)^2 - \frac{1}{2} \right) \) clearly diverges since \( (\ln t)^2 \to \infty \). This divergence tells us that the original integral does not settle to a specific value but rather extends indefinitely, highlighting its divergence.
Natural Logarithm Integration
Integrating functions involving the natural logarithm function, \( \ln x \), can be unique and enriching. In our problem, we tackle a specific function involving \( \ln x \):
- Here, \( \frac{\ln x}{x} \) is integrated over the interval from \( e \) to \( t \).
- The clever insight is to recognize that \( \frac{\ln x}{x} \) is the derivative of \( \frac{1}{2} (\ln x)^2 \).
- Integrating, we obtain \( \int \frac{\ln x}{x} \, dx = \frac{1}{2}(\ln x)^2 + C \).