Math Problem Statement

Evaluate the integral ∫ ln(1/x) dx over the specified bounds.

Solution

The integral provided in the image is:

0ln(1x)dx.\int_{0}^{\infty} \ln\left(\frac{1}{x}\right) dx.

Let me evaluate this integral step-by-step.


Step 1: Simplify the logarithmic term

The logarithmic property ln(1x)=ln(x)\ln\left(\frac{1}{x}\right) = -\ln(x) can be applied. Thus, the integral becomes: 0ln(1x)dx=0ln(x)dx.\int_{0}^{\infty} \ln\left(\frac{1}{x}\right) dx = -\int_{0}^{\infty} \ln(x) dx.


Step 2: Analyze the bounds of the integral

The integral 0ln(x)dx-\int_{0}^{\infty} \ln(x) dx is improper, requiring further evaluation because the function ln(x)\ln(x) is undefined at x=0x = 0 and grows unbounded as xx \to \infty.


Step 3: Consider convergence

  • Near x=0x = 0, ln(x)\ln(x) diverges to -\infty.
  • As xx \to \infty, ln(x)\ln(x) diverges to ++\infty.

Thus, the integral does not converge, and the result is divergent.


If you have any additional questions or wish for a more detailed explanation of the divergence, feel free to ask!


Related Questions:

  1. How do you determine if an improper integral converges or diverges?
  2. What is the difference between logarithmic and exponential divergence?
  3. Could a substitution like u=ln(x)u = \ln(x) help analyze a similar integral?
  4. Are there specific conditions under which 0f(x)dx\int_{0}^{\infty} f(x) \, dx converges?
  5. How does the logarithm function behave as x0x \to 0 and xx \to \infty?

Tip:

When solving improper integrals, always check for convergence before proceeding with formal calculations.

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Math Problem Analysis

Mathematical Concepts

Calculus
Improper Integrals
Logarithmic Functions

Formulas

ln(1/x) = -ln(x)

Theorems

Convergence and Divergence of Improper Integrals

Suitable Grade Level

Undergraduate - Calculus I or II