Problem 57

Question

For what positive values of $p$ is the improper integral $\int_{1}^{e} \frac{1}{x \ln ^{p}(x)} d x$ convergent?

Step-by-Step Solution

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Answer

The integral converges for positive values of $ p $ where $ 0 < p < 1 $.

1Step 1: Identify the Type of Integral

The given integral is improper because its integrand becomes infinite at the lower limit, as $ x \to 1^+ $, where $ \ln(x) \to 0 $. We need to determine when this improper integral converges.

2Step 2: Simplify the Problem with Substitution

Let's perform a substitution to simplify the integrand. Set $ u = \ln(x) $, which means $ du = \frac{1}{x} dx $. When $ x = 1 $, $ u = 0 $ and when $ x = e $, $ u = 1 $. The integral becomes $ \int_{0}^{1} \frac{1}{u^p} du $.

3Step 3: Determine Conditions for Convergence

The integral $ \int_{0}^{1} \frac{1}{u^p} du $ is now in a standard form. This integral converges if $ 1 - p > 0 $, which simplifies to $ p < 1 $. This is because $ \int_{0}^{1} u^{-p} du $ converges for $ u^q $ when $ q > -1 $.

4Step 4: Conclude the Range of $ p $

The improper integral $ \int_{1}^{e} \frac{1}{x \ln^p(x)} dx $ converges for $ 0 < p < 1 $. If $ p \geq 1 $, the integral diverges because the power of $ u $ would not satisfy the convergence condition.

Key Concepts

Substitution MethodConvergence CriteriaLimits of Integration

Substitution Method

To tackle challenging integrals, particularly improper ones, the substitution method is a powerful tool. It simplifies the original problem, allowing us to work with a more manageable expression.

In our exercise, dealing with the integral $ \int_{1}^{e} \frac{1}{x \ln^p(x)} dx $, we made a smart substitution by setting $ u = \ln(x) $. This effectively transformed our integrand, with the derivative $ du = \frac{1}{x} dx $.

Initially, when $ x = 1 $, $ u = 0 $.
When $ x = e $, $ u = 1 $.

This resulted in a new integral: $ \int_{0}^{1} \frac{1}{u^p} du $, which is much easier to evaluate. The substitution method helps by converting to variables or expressions that simplify integration limits and integrands. This method is central in proving convergence in improper integrals, as it simplifies the understanding of complex expressions.

Convergence Criteria

Convergence in integrals refers to the behavior of the integral as it approaches its bounds. The integral will "converge" when it settles on a specific value.

For the transformed integral $ \int_{0}^{1} \frac{1}{u^p} du $, convergence depends on the power of $ u $. This is determined by the convergence condition concerning the exponent in the integrand:

Our integral converges if $ 1 - p > 0 $, which simplifies to $ p < 1 $.

Intrinsically, this requires that the expression inside the integral remains manageable near the limits of integration. A fundamental aspect of evaluating convergence is ensuring the integrated function's exponents align with known convergent integral forms, like $ \int_{0}^{1} u^q du $, which converges when $ q > -1 $.

Thus, recognizing such critical thresholds for $ p $ allows us to determine convergence accurately.

Limits of Integration

The limits of integration in an integral can heavily influence its convergence or divergence. These limits mark the start and end points over which the function is evaluated.

In the integral $ \int_{1}^{e} \frac{1}{x \ln^p(x)} dx $, the limits of integration are $ 1 $ and $ e $, the base of natural logarithms. When substituted, these translate to limits $ 0 $ to $ 1 $ for $ u $.

Understanding these limits is crucial, as the behavior of the integrand near these bounds determines the integral's nature:

At $ u = 0 $ (formerly $ x=1 $), $ \ln(x) \to 0 $ makes the integral improper because the function may "blow up" as it approaches an infinite limit.
At $ u = 1 $ (formerly $ x=e $), the expression is well-behaved and bounded.

In this specific case, $ u=0 $ is where one must examine convergence issues, as it provides insight into how the integrand behaves as it approaches potentially hazardous bounds. The accurate transformation of limits helps us judge the behavior of integrals around critical points, ensuring a comprehensive evaluation of improper integrals.

Problem 56

Problem 57

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