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Use L'Hôpital's rule to help evaluate the improper integral. $$\int_{1}^{+\infty} \frac{\ln x}{x^{2}} d x$$

Short Answer

Expert verified
The integral converges to -1.

Step by step solution

01

Identify the Type of Integral

The given integral \( \int_{1}^{+\infty} \frac{\ln x}{x^2} \, dx \) is an improper integral because of the infinite upper limit. We need to evaluate this integral using limits.
02

Set Up the Limit for the Improper Integral

Convert the improper integral to a limit by replacing the upper bound \(+\infty\) with a variable \(b\) and then taking the limit as \(b\) approaches infinity:\[\lim_{b \to +\infty} \int_{1}^{b} \frac{\ln x}{x^2} \, dx.\]
03

Integration by Parts Setup

Use integration by parts where \(u = \ln x\) and \(dv = \frac{1}{x^2} \, dx\). Then, \(du = \frac{1}{x} \, dx\) and \(v = -\frac{1}{x}\).
04

Apply Integration by Parts

Using the integration by parts formula \(\int u \, dv = uv - \int v \, du\), compute the integral:\[\int \ln x \cdot \frac{1}{x^2} \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx.\]This simplifies to:\[-\frac{\ln x}{x} - \frac{1}{x} + C.\]
05

Evaluate the Antiderivative from 1 to b

Substitute the evaluated antiderivative at the bounds 1 and \(b\):\[\left[-\frac{\ln x}{x} - \frac{1}{x}\right]_{1}^{b} = \left(-\frac{\ln b}{b} - \frac{1}{b}\right) - \left(-\frac{\ln 1}{1} - \frac{1}{1}\right).\]This simplifies to:\[\left(-\frac{\ln b}{b} - \frac{1}{b}\right) - (0 + 1).\]
06

Apply the Limit as \(b\) Approaches Infinity

Evaluate the limit:\[\lim_{b \to +\infty} \left(-\frac{\ln b}{b} - \frac{1}{b} - 1\right).\]Since both \(\frac{\ln b}{b}\) and \(\frac{1}{b}\) approach 0 as \(b\) approaches infinity, the limit simplifies to:\[-1.\]
07

Conclude the Evaluation

Conclude that the value of the improper integral is \(-1\). This means the integral \(\int_{1}^{+\infty} \frac{\ln x}{x^2} \, dx\) converges to \(-1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

L'Hôpital's rule
L'Hôpital's rule is an essential tool in calculus used to evaluate limits that involve indeterminate forms, typically in the form of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with such limits, we can apply L'Hôpital's rule by differentiating the numerator and the denominator separately and then taking the limit again.
This rule simplifies finding limits where straightforward substitution does not work.
  • Identify when an indeterminate form is present.
  • Differentiate both the numerator and the denominator.
  • Evaluate the limit of the new function formed.
Remember, L'Hôpital's rule can be applied iteratively if the first application still results in an indeterminate form. However, it is crucial that the conditions for L'Hôpital's rule are satisfied: the original form must be indeterminate and the derivatives of the functions in question must exist.
Integration by parts
Integration by parts is a technique derived from the product rule of differentiation, used for integrating products of functions. The formula is given by \( \int u \, dv = uv - \int v \, du \), where \( u \) and \( v \) are differentiable functions.
Choosing \( u \) and \( dv \) correctly can simplify the integration process significantly. Here are the basic steps involved:
  • Choose \( u \) and \( dv \) such that \( du \) and \( v \) are easy to compute.
  • Compute \( du = (du/dx) \, dx \) and \( v = \int dv \).
  • Substitute back into the formula \( \int u \, dv = uv - \int v \, du \).
Apply integration by parts iteratively if necessary until the integral is fully evaluated. This technique is particularly useful with logarithmic functions, polynomials, and exponentials when they appear together as products in integrals.
Limits
Limits are a core concept in calculus that describe the behavior of a function as its argument approaches a certain point. In the context of improper integrals, limits are used to handle the infinite bounds or discontinuities.
When evaluating an improper integral, replace the infinite bound with a variable and then compute the limit as this variable approaches infinity or the point of discontinuity.
  • Transform the integral by replacing the problematic bound with a variable \( b \).
  • Compute the integral with the variable instead of the infinity bound.
  • Take the limit as the variable approaches the bound (infinity, in this case).
If the limit exists, the integral is said to converge. If the limit does not exist or is infinite, the integral diverges. Limits are fundamental in determining the convergence or divergence of improper integrals.
Convergence of integrals
Convergence of integrals concerns whether an integral approaches a finite value as one or more of its boundaries approach infinity or a singularity. An improper integral converges if, as the variable bound approaches the limit, the integral approaches a finite number.
In determining convergence, certain criteria and techniques can be helpful:
  • Simplify the integral using substitution or integration techniques like integration by parts.
  • Use limits to replace infinite bounds or discontinuities and evaluate the resulting expression.
  • Apply tests for convergence, such as the comparison test, to determine if an integral converges or diverges.
Properly evaluating convergence ensures that the calculated integral has a meaningful interpretation. Understanding when an integral converges is critical for dealing with real-world applications, especially in physics and engineering scenarios.

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