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Chapter 7: Problem 68

Use integration by parts to evaluate the following integrals. $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$$

Short Answer

Expert verified
Answer: The value of the integral is $$1$$.

Step by step solution

01

Identify the functions to be differentiated and integrated in integration by parts

We have the given integral: $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x$$ We will use integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ Let's choose \(u = \ln x\) and \(dv = \frac{1}{x^2} dx\).
02

Differentiate u and integrate dv

Now we find the derivative of \(u\) and the integral of \(dv\): $$du=\frac{d(\ln x)}{dx}dx=\frac{1}{x} dx$$ $$v=\int \frac{1}{x^{2}} d x = - \frac{1}{x}$$
03

Apply integration by parts

Substitute the result from step 2 into the integration by parts formula: $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x = \left[-\frac{\ln x}{x}\right]_{1}^{\infty} - \int_{1}^{\infty} \frac{-1}{x^2} dx$$
04

Evaluate the remaining integral

The remaining integral can be easily evaluated: $$-\int_{1}^{\infty} \frac{-1}{x^2} dx = \int_{1}^{\infty} \frac{1}{x^2}dx = \left[- \frac{1}{x}\right]_{1}^{\infty}$$
05

Evaluate the limits

Now, we will evaluate the limits for both the functions: As \(x \to \infty\), $$\frac{\ln x}{x} \to 0$$ $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x = \left[-\frac{\ln x}{x}\right]_{1}^{\infty} - \left[- \frac{1}{x}\right]_{1}^{\infty} = 0 - (-1+0) = 1$$ So, the final answer is: $$\int_{1}^{\infty} \frac{\ln x}{x^{2}} d x = 1$$

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Key Concepts

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

Improper Integrals
Improper integrals are a type of integral where either the interval of integration is infinite or the integrand becomes infinitely large at one or more points within the interval of integration. In this context, the integral we are dealing with is \[ \int_{1}^{\infty} \frac{\ln x}{x^{2}} d x \] and it is classified as improper because the upper limit of integration extends to infinity. When evaluating an improper integral, it's crucial to assess if the integral converges (results in a finite value).
  • Convergence and Divergence: If the limit exists and is finite as you approach the endpoint(s), the integral converges. Otherwise, it diverges.
  • Evaluating: For an infinite interval, replace the infinity with a variable (e.g., \( b \)), and consider the limit as \( b \to \infty \).
Understanding how to handle these limits is key to solving improper integrals, ensuring that the solution is both correct and meaningful.
Logarithmic Functions
Logarithmic functions, such as \( \ln x \), play an essential role in calculus, particularly in integration via certain techniques like integration by parts. The natural logarithm \( \ln x \) is differentiable for all \( x > 0 \), which is useful in various integration techniques.
Here are some key points about logarithmic functions:
  • Derivative: The derivative of \( \ln x \) is \( \frac{1}{x} \), a simple yet powerful derivative that's often used in integration by parts to simplify problems involving products.
  • Properties: Understanding properties like \( \ln(ab) = \ln a + \ln b \) and \( \ln(a^b) = b\ln a \) can help simplify problems during integration.
  • Applications: Logarithmic functions appear naturally in problems involving exponential growth and decay, making them vital for real-world applications.
When dealing with logarithmic functions in integration, carefully choosing \( u \) and \( dv \) when applying techniques, such as integration by parts, can significantly simplify the process.
Calculus Integration Techniques
Integration is a central concept in calculus, involving finding the antiderivative of a function. Various techniques exist to handle different forms of integrals effectively, such as substitution, partial fraction decomposition, and integration by parts.
  • Integration by Parts: This method is based on the product rule for differentiation and is particularly useful for integrals involving a product of functions. It states:\[ \int u \, dv = uv - \int v \, du \]
  • Application: Choosing which part of the integrand to differentiate and which to integrate is crucial. Typically, functions like logarithms or algebraic expressions become \( u \), while \( dv \) is an exponential or trigonometric function that simplifies on integration.
In our problem, \( u = \ln x \) was chosen because its derivative simplifies the expression, helping solve the integral. By mastering integration techniques, one can tackle a wide array of integrals, enhancing both theoretical understanding and practical problem-solving skills.

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Most popular questions from this chapter

The Eiffel Tower property Let \(R\) be the region between the curves \(y=e^{-\alpha x}\) and \(y=-e^{-\alpha x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c>0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves; see the Guided Project The exponential Eiffel Tower.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\).

Recall that the substitution \(x=a \sec \theta\) implies either \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0)\) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). Graph the function \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\) and consider the region bounded by the curve and the \(x\) -axis on \([-6,-3] .\) Then evaluate \(\int_{-6}^{-3} \frac{\sqrt{x^{2}-9}}{x} d x .\) Be sure the result is consistent with the graph.

Evaluate the following integrals. $$\int_{0}^{a} x^{x}(\ln x+1) d x, a>0$$

A recent 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

Many methods needed Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x\) b. Change variables by letting \(y=1 / x\) c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) (and that both integrals converge). Conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x-0\) d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\)
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