Evaluating The Definite Integral Of 1/(x Ln(x^2)) From 1 To 2

Introduction

In this article, we will delve into the evaluation of a specific definite integral, a quintessential problem in calculus. Our focus is on the integral of the function ${ \frac{1}{x \ln(x^2)} }$ with respect to ${ x }$, over the interval from 1 to 2. Definite integrals, which represent the signed area under a curve between two points, are a cornerstone of calculus and find extensive applications in various fields, including physics, engineering, and economics. The function at hand presents an interesting challenge due to the presence of both a reciprocal term and a logarithmic function in the denominator. To solve this, we will employ a combination of techniques, including u-substitution and the properties of logarithms, to simplify the integral and ultimately find its exact value. This exploration is not just an exercise in mathematical manipulation but also an opportunity to reinforce fundamental concepts in integral calculus and appreciate the elegance of mathematical problem-solving. Understanding how to tackle such integrals is crucial for anyone studying calculus or related disciplines, as it showcases the power and versatility of integration techniques in handling complex functions.

Problem Statement

Our task is to evaluate the definite integral:

${ \int_{1}^{2} \frac{1}{x \ln(x^2)} dx }$

This problem involves integrating a function that combines algebraic and logarithmic elements, requiring a strategic approach to simplify and solve. The presence of ${ x }$ in the denominator, coupled with the natural logarithm of ${ x^2 }$, suggests that a suitable substitution might be beneficial. The limits of integration, from 1 to 2, define the interval over which we are calculating the signed area under the curve of the given function. To proceed, we will first explore simplification techniques, focusing on leveraging properties of logarithms and identifying a viable substitution that transforms the integral into a more manageable form. The goal is to find an antiderivative of the function and then apply the Fundamental Theorem of Calculus to evaluate the definite integral over the specified interval. This process demonstrates the practical application of calculus in determining exact values for integrals involving complex functions, a skill vital in numerous scientific and engineering contexts. By systematically breaking down the problem, we aim to provide a clear and concise solution that illuminates the underlying mathematical principles and techniques.

Simplification Using Logarithmic Properties

Before we jump into integration techniques, let’s simplify the integrand using logarithmic properties. Recall that ${ \ln(a^b) = b \ln(a) }$. Applying this property to our integral, we get:

${ \ln(x^2) = 2 \ln(x) }$

Thus, the integral becomes:

${ \int_{1}^{2} \frac{1}{x \cdot 2 \ln(x)} dx = \frac{1}{2} \int_{1}^{2} \frac{1}{x \ln(x)} dx }$

This simplification is a crucial step in making the integral more tractable. By reducing ${ \ln(x^2) }$ to ${ 2 \ln(x) }$, we've eliminated an unnecessary layer of complexity. The factor of 2 in the denominator can be easily factored out as ${ \frac{1}{2} }$, leaving us with a simpler integral to solve. The simplified integrand, ${ \frac{1}{x \ln(x)} }$, now exhibits a structure that is more amenable to standard integration techniques, particularly u-substitution. Recognizing and applying logarithmic properties like this is a fundamental skill in calculus, allowing us to transform complex expressions into simpler forms that are easier to manipulate. This step underscores the importance of algebraic proficiency in tackling calculus problems, as simplification often paves the way for successful integration. The revised integral is now primed for a substitution that will further streamline the integration process and lead us toward the final solution. This initial simplification highlights how strategic manipulation can significantly reduce the difficulty of integration problems.

Applying u-Substitution

To evaluate the simplified integral, we’ll use u-substitution. Let:

${ u = \ln(x)}$

Then, the derivative of ${ u }$ with respect to ${ x }$ is:

${ \frac{du}{dx} = \frac{1}{x} }$

Which implies:

${ du = \frac{1}{x} dx }$

Now, we need to change the limits of integration according to our substitution. When ${ x = 1 }$, ${ u = \ln(1) = 0 }$, and when ${ x = 2 }$, ${ u = \ln(2) }$. Thus, our integral transforms to:

${ \frac{1}{2} \int_{0}^{\ln(2)} \frac{1}{u} du }$

The u-substitution is a powerful technique that simplifies integrals by replacing a complex expression with a single variable. In this case, substituting ${ u = \ln(x) }$ effectively transforms the integral into a form that is readily integrable. The key to a successful u-substitution lies in choosing the right expression for ${ u }$ such that its derivative appears in the original integral, allowing for a clean substitution. The change in limits of integration is also a crucial step, ensuring that the definite integral is evaluated correctly in terms of the new variable. By transforming the integral into ${ \frac{1}{2} \int_{0}^{\ln(2)} \frac{1}{u} du }$, we have significantly simplified the problem. This new integral is a basic form that we can easily integrate using the known antiderivative of ${ \frac{1}{u} }$, which is ${ \ln|u| }$. This step showcases the elegance and efficiency of u-substitution in handling integrals involving composite functions, making it an indispensable tool in integral calculus.

Evaluating the Integral

The integral ${ \int \frac{1}{u} du }$ is a standard integral with the antiderivative ${ \ln|u| }$. Therefore, we have:

${ \frac{1}{2} \int_{0}^{\ln(2)} \frac{1}{u} du = \frac{1}{2} [\ln|u|]_{0}^{\ln(2)} }$

Now, we evaluate the antiderivative at the limits of integration:

${ \frac{1}{2} [\ln(\ln(2)) - \lim_{u \to 0^+} \ln(u)] }$

As ${ u }$ approaches 0 from the positive side, ${ \ln(u) }$ approaches negative infinity. This indicates that the integral is improper.

To properly evaluate this improper integral, we need to consider the limit:

${ \lim_{a \to 0^+} \frac{1}{2} \int_{a}^{\ln(2)} \frac{1}{u} du = \lim_{a \to 0^+} \frac{1}{2} [\ln(u)]_{a}^{\ln(2)} }$

${ = \frac{1}{2} \lim_{a \to 0^+} [\ln(\ln(2)) - \ln(a)] }$

Since ${ \lim_{a \to 0^+} \ln(a) = -\infty }$, the limit becomes:

${ \frac{1}{2} [\ln(\ln(2)) - (-\infty)] = \infty }$

Thus, the integral diverges.

Evaluating the integral involves applying the Fundamental Theorem of Calculus, which states that the definite integral of a function can be found by evaluating its antiderivative at the upper and lower limits of integration. In this case, the antiderivative of ${ \frac{1}{u} }$ is ${ \ln|u| }$. However, when we substitute the limits of integration, we encounter a situation where the natural logarithm of 0 is undefined, indicating an improper integral. To handle this, we introduce a limit as the lower bound approaches 0 from the positive side. This process reveals that the integral diverges, meaning it does not have a finite value. The divergence arises because the function ${ \frac{1}{u} }$ has a singularity at ${ u = 0 }$, and the area under the curve becomes infinitely large as we approach this singularity. Understanding how to handle improper integrals is a critical skill in calculus, as many real-world applications involve functions with singularities or infinite limits of integration. This evaluation underscores the importance of careful analysis when dealing with integrals and the potential for seemingly straightforward integrals to exhibit divergent behavior.

Conclusion

In conclusion, we set out to evaluate the definite integral:

${ \int_{1}^{2} \frac{1}{x \ln(x^2)} dx }$

Through simplification using logarithmic properties and employing u-substitution, we transformed the integral into:

${ \frac{1}{2} \int_{0}^{\ln(2)} \frac{1}{u} du }$

However, upon evaluating this integral, we found that it diverges due to the lower limit of integration approaching 0, where the natural logarithm function tends to negative infinity. This divergence indicates that the area under the curve of the function ${ \frac{1}{x \ln(x^2)} }$ from 1 to 2 is unbounded. The process of evaluating this integral highlights several key concepts in calculus. First, the importance of simplifying the integrand using properties of logarithms to make it more manageable. Second, the effectiveness of u-substitution in transforming integrals into simpler forms. Third, the recognition and handling of improper integrals, which require special techniques to evaluate due to singularities or infinite limits of integration. This problem serves as a valuable example of how seemingly straightforward integrals can lead to complex analyses and the necessity of understanding the behavior of functions near singularities. The result, that the integral diverges, is a significant finding that underscores the importance of rigorous evaluation techniques in calculus.