Question

Evaluate each improper integral or show that it diverges. \(\int_{2}^{\infty} \frac{\ln x}{x^{2}} d x\)

Short answer

The integral converges to \(\frac{\ln 2}{2} + \frac{1}{2}\).

Step-by-step solution

Understand the Problem

We need to evaluate the improper integral \(\int_{2}^{\infty} \frac{\ln x}{x^{2}} dx\) or demonstrate that it diverges. This integral is improper because the upper limit is infinite.

Rewrite the Integral with a Limit

Since the integral is improper due to the infinite bound, we replace the infinity with a variable \(b\) and take the limit as \(b\) approaches infinity: \[\lim_{b \to \infty} \int_{2}^{b} \frac{\ln x}{x^{2}} dx.\]

Integration by Parts

To evaluate the integral, we use integration by parts. Let \(u = \ln x\) and \(dv = \frac{1}{x^2} dx\). Then, \(du = \frac{1}{x} dx\) and \(v = -\frac{1}{x}\). Applying integration by parts, \[\int u \, dv = uv - \int v \, du.\]Substitute the values:\[-\frac{\ln x}{x} \bigg|_{2}^{b} - \int_{2}^{b} \left(-\frac{1}{x}\right) \frac{1}{x} \, dx.\]

Simplify the Expression

The expression becomes:\[-\frac{\ln x}{x} \bigg|_{2}^{b} + \int_{2}^{b} \frac{1}{x^2} dx.\]Evaluate \(\int \frac{1}{x^2} dx\):\[\int \frac{1}{x^2} dx = -\frac{1}{x} \bigg|_{2}^{b}.\]

Evaluate the Limits

Substitute the evaluated integral back:\[-\frac{\ln x}{x} \bigg|_{2}^{b} - \frac{1}{x} \bigg|_{2}^{b} = \left(-\frac{\ln b}{b} + \frac{\ln 2}{2}\right) + \left(-\frac{1}{b} + \frac{1}{2}\right).\]As \(b \to \infty\), both \(-\frac{\ln b}{b}\) and \(-\frac{1}{b}\) approach 0. Thus, the limit of the expression is \(\frac{\ln 2}{2} + \frac{1}{2}\).

Conclusion

Combine constants: \[\frac{\ln 2}{2} + \frac{1}{2}.\] The integral converges to a finite limit.

Key concepts

Integration by Parts

When faced with complex integrals, one of the most useful tools at our disposal in calculus is the technique of integration by parts. This method is particularly handy when dealing with the integral of a product of functions. To use integration by parts, we identify two components within the integrand: one that will be our "u" (which we differentiate), and another, "dv" (which we integrate). The formula for integration by parts stems from the product rule for derivatives and is given by:

• \[ \int u \, dv = uv - \int v \, du \]

In our problem, we identified \( u = \ln x \) and \( dv = \frac{1}{x^2} \, dx \). Differentiating \( u \), we find \( du = \frac{1}{x} \, dx \); integrating \( dv \), we get \( v = -\frac{1}{x} \). After substitution, this allows us to break down the original integral into more manageable parts.
Using this technique helps transform complex expressions into simpler ones that are easier to integrate, which is crucial for solving improper integrals like the one we had in our exercise.

Convergence of Integrals

In the realm of improper integrals, determining whether an integral converges or diverges is a fundamental task. A convergent integral provides a finite value as its result, while a divergent integral does not settle on a particular number.
To evaluate the convergence of the integral \( \int_{2}^{\infty} \frac{\ln x}{x^{2}} \, dx \), we substitute the infinite limit with a variable \( b \) and take the limit as \( b \) approaches infinity:

• \[ \lim_{b \to \infty} \int_{2}^{b} \frac{\ln x}{x^{2}} \, dx \]

Through the evaluation steps, we notice that as \( b \to \infty \), terms like \(-\frac{\ln b}{b}\) and \(-\frac{1}{b}\) approach zero. Thus, the resulting finite value reflects the behavior of the entire integral.
Recognizing when such an integral is convergent often involves checking how the integrand behaves as approaches its infinite limit. In this exercise, as \( b \) grew indefinitely large, both crucial terms diminished to zero, indicating convergence.

Calculus Techniques

The effective application of different calculus techniques is essential for tackling improper integrals. In the solved exercise, we employed both integration by parts and the properties of limits to reach a solution.
Here's a quick overview of the procedures followed:

• Switching an infinite bound to a variable \( b \) and introducing the limit allows us to handle the improper integral.
• Applying integration by parts helps decompose the integral into simpler terms that are more straightforward to integrate.
• This process requires careful handling of logarithmic and polynomial terms, which often appear in these integrals.
• Finally, evaluating the limit as \( b \to \infty \) confirms the convergence of the integral.

These techniques demonstrate how calculus provides a versatile toolkit for evaluating otherwise indeterminate forms, guiding us smoothly through complex problems.

Infinite Limits

Working with infinite limits can appear intimidating, but breaking down each step can make the process digestible. In the context of improper integrals, infinite limits help us evaluate how integrands behave as they theoretically stretch beyond bounds.
In the specific case of the integral \( \int_{2}^{\infty} \frac{\ln x}{x^{2}} \, dx \), replacing infinity with a finite limit \( b \) offers us a temporary relief from endlessness. By calculating the limit

• \[ \lim_{b \to \infty} \int_{2}^{b} \frac{\ln x}{x^{2}} \, dx \]

we observe how the expression evolves. During the solution, examining terms like \(-\frac{\ln b}{b}\) helps us understand the decay of influence as \( b \) increases. An intuitive comprehension of this diminishing influence helps affirm that the integral, eventually, converges.
This understanding of infinite limits shows their practicality in evaluating the behavior of functions extending to infinity, a critical part of solving improper integrals.