Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Determine whether the following integrals converge or diverge. $$\int_{3}^{\infty} \frac{d x}{\ln x}(\text { Hint: } \ln x \leq x .)$$

Short Answer

Expert verified
Answer: The improper integral \(\int_{3}^{\infty} \frac{dx}{\ln x}\) diverges.

Step by step solution

01

Identify the integral as improper

We can see that the given integral is improper because it has an infinite limit \(\infty\) in the upper bound. In order to determine if it converges or diverges, we will use the comparison test for integrals.
02

Use the comparison test

According to the comparison test for integrals, given two functions \(f(x)\) and \(g(x)\), if \(0 \leq f(x) \leq g(x)\) for all x in the interval \([a, \infty)\), then: - if \(\int_a^{\infty} g(x)dx\) converges, then \(\int_a^{\infty} f(x)dx\) also converges, - if \(\int_a^{\infty} f(x)dx\) diverges, then \(\int_a^{\infty} g(x)dx\) also diverges. We're given the hint that \(\ln x \leq x\), so we can use it to find a function \(g(x)\) to compare with the given integral.
03

Compare with a simpler function

Using the hint, we have: $$\frac{1}{\ln x} \geq \frac{1}{x}$$ Now we can set \(f(x) = \frac{1}{\ln x}\) and \(g(x) = \frac{1}{x}\). Then, our inequality becomes: $$0 \leq f(x) = \frac{1}{\ln x} \geq g(x) = \frac{1}{x}$$ for all \(x\) in the interval \([3, \infty)\).
04

Check if the integral of the simpler function converges or diverges

First, let's integrate the function \(g(x)\): $$\int_3^{\infty} \frac{1}{x} dx$$ To solve this improper integral, we rewrite it as a limit: $$\lim_{t\to\infty} \int_3^{t} \frac{1}{x} dx$$ Now, integrate \(g(x) = \frac{1}{x}\): $$\lim_{t\to\infty}\left[\ln|x|\right]_3^t$$ Now evaluate the limit: $$\lim_{t\to\infty} (\ln|t| - \ln|3|) = \infty$$ Since the integral of the simpler function \(g(x)\) diverges, we can conclude that our initial integral also diverges according to the comparison test.
05

Conclude the result

By applying the comparison test for integrals, using the hint provided, and detecting the divergence of the simpler function, we can conclude that the improper integral \(\int_{3}^{\infty} \frac{dx}{\ln x}\) diverges.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Comparison Test
The Comparison Test is a powerful tool used to determine whether an improper integral converges or diverges by comparing it with another integral whose behavior is already known.
It is especially helpful when working with functions that are difficult to integrate directly.
  • We say we have two functions, \( f(x) \) and \( g(x) \).
  • If \( g(x) \) is larger or equal to \( f(x) \) over a certain interval \([a, \infty)\), and the integral \( \int_a^{\infty} g(x) \, dx \) converges, then the integral \( \int_a^{\infty} f(x) \, dx \) also converges.
  • Conversely, if \( \int_a^{\infty} f(x) \, dx \) diverges, then \( \int_a^{\infty} g(x) \, dx \) also diverges.
Utilizing this method requires identifying appropriate comparison functions and ensuring their relationships are valid over the specified interval.
Divergence
Divergence in the context of improper integrals refers to the situation where the integral does not settle to a finite number as one or both limits of integration approach infinity or extend to a discontinuity.
This concept helps us understand when an integral cannot be evaluated to a finite value.
  • An integral \( \int_a^{\infty} f(x) \, dx \) is said to diverge if, as we take the limit of the upper bound to infinity, the integral keeps increasing without bound.
  • In the case of \( \frac{1}{x} \), as seen in the solution, the indefinite integral leads to \( \lim_{t\to\infty} (\ln|t| - \ln|a|) = \infty \). This indicates divergence because the result tends towards infinity rather than stabilizing.
Recognizing divergence is crucial in determining the behavior of improper integrals, which assists in applying the comparison test accurately.
Limit of Integration
The Limit of Integration is a vital concept when dealing with improper integrals.
It highlights how the boundaries of integration—especially when they involve infinity—affect the nature of the solutions we obtain.
  • Improper integrals arise when one or both limits extend to infinity, or when the function is undefined at a point within the limits.
  • The process involves taking a limit to substitute the infinite bound—e.g., converting \( \int_a^{\infty} f(x) \, dx \) to \( \lim_{t\to \infty} \int_a^{t} f(x) \, dx \).
  • This approach helps in handling the tension of infinite intervals, allowing us to examine whether integration results in finite or infinite values, which is crucial in determining convergence or divergence.
Understanding limits of integration is foundational for making well-informed decisions about the behavior of improper integrals.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Normal distribution of movie lengths A study revealed that the lengths of U.S. movies are normally distributed with a mean of 110 minutes and a standard deviation of 22 minutes. This means that the fraction of movies with lengths between \(a\) and \(b\) minutes (with \(a

Estimating error Refer to Theorem 8.1 in the following exercises. Let \(f(x)=\cos x^{2}\) a. Find a Midpoint Rule approximation to \(\int_{-1}^{1} \cos x^{2} d x\) using \(n=30\) subintervals. b. Calculate \(f^{\prime \prime}(x)\) c. Explain why \(\left|f^{*}(x)\right| \leq 6\) on [-1,1] d. Use Theorem 8.1 to find an upper bound on the absolute error in the estimate found in part (a).

The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system. $$\int \frac{\sin ^{-1} a x}{x^{2}} d x, a>0$$

A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated using integration by parts: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=1 \rightarrow F(s)=\frac{1}{s}$$

Trapezoid Rule and Simpson's Rule Consider the following integrals and the given values of \(n .\) a. Find the Trapezoid Rule approximations to the integral using \(n\) and \(2 n\) subintervals. b. Find the Simpson's Rule approximation to the integral using \(2 n\) subintervals. It is easiest to obtain Simpson's Rule approximations from the Trapezoid Rule approximations, as in Example \(8 .\) c. Compute the absolute errors in the Trapezoid Rule and Simpson's Rule with \(2 n\) subintervals. $$\int_{0}^{\pi / 4} \frac{d x}{1+x^{2}} ; n=64$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free