Problem 89 - Thomas Calculus

Question

In Exercises \(89-92\) , use a CAS to explore the integrals for various values of \(p\) (include noninteger values). For what values of \(p\) does the integral converge? What is the value of the integral when it does converge? Plot the integrand for various values of \(p .\) $$ \int_{0}^{e} x^{p} \ln x d x $$

Step-by-Step Solution

Answer

The integral converges for \( p > 0 \) and for these \( p \) values, the integral value depends on specific \( p \) (e.g., \( \frac{e^2}{4} \) for \( p = 1 \)).

Step 1: Analyze the Problem

We need to evaluate the integral \( \int_{0}^{e} x^{p} \ln x \, dx \) and determine for which values of \( p \) the integral converges. Exploring convergence involves checking the behavior of the integrand at the endpoints \( x = 0 \) and \( x = e \).

Step 2: Consider Convergence at Lower Limit

First, analyze the behavior at the lower limit, \( x = 0 \). For small \( x \), \( \ln x \) approaches \(-\infty\). The integrand behaves like \( x^p \ln x \approx x^p (-\infty) \), which suggests divergence unless \( p > 0 \) where it converges because \( x^p \) approaches \( 0 \) faster than \( \ln x \) diverges.

Step 3: Consider Convergence at Upper Limit

At the upper limit, \( x = e \), \( \ln x = 1 \), so the integrand \( x^p \ln x \approx e^p \). Since \( e^p \) is finite for any \( p \), the integral converges at the upper limit.

Step 4: Conclude on Convergence

Combining both endpoints, the integral converges for \( p > 0 \). For \( p \leq 0 \), the integral involves terms that diverge at the lower limit (especially for nonpositive \( p \)).

Step 5: Calculate the Integral

Using integration by parts (choose \( u = \ln x \), \( dv = x^p \, dx \)), it helps to evaluate the integral analytically. For convergent \( p \) values, compute it to confirm results. For example, when \( p = 1 \), \[ \int_{0}^{e} x \ln x \, dx = \left. \frac{x^2}{2} \ln x \right|_{0}^{e} - \int_{0}^{e} \frac{x^2}{2x} \, dx = \frac{e^2}{2} - \left. \frac{x^2}{4} \right|_{0}^{e} = \frac{e^2}{4}. \]

Step 6: Plot the Integrand

Plot \( x^p \ln x \) for several values of \( p \), especially for some positive and negative values to visualize convergence behavior. For example, for \( p = 1/2, 0, -1/2 \), plot from \( 0 \) to \( e \) to illustrate how increasing \( p \) influences convergence.

Key Concepts

Convergence of Integrals

The convergence of an integral essentially indicates if its value approaches a finite number as the limits of integration are reached. In our case, we are working with the integral \( \int_{0}^{e} x^{p} \ln x \, dx \). Determining when this integral converges is essential to finding its value.

To analyze convergence, we look at the behavior of the integrand \( x^{p} \ln x \) at the endpoints, \( x = 0 \) and \( x = e \). At the lower end, unlike \( x^{p} \), the \( \ln x \) approaches \(-\infty\) as \( x \) approaches 0. For the integral to converge, \( x^{p} \) must approach 0 faster than \( \ln x \) diverges. This occurs when \( p > 0 \).

At the upper limit, \( x = e \), the behavior is simpler, as \( \ln x = 1 \). This makes the integrand behave like \( e^{p} \), which remains finite. Thus, the overall conclusion is that the integral converges when \( p > 0 \), as both endpoints are handled appropriately in this case.

Integration by Parts

Integration by parts is a powerful technique for evaluating integrals, especially when dealing with products of functions, such as \( x^{p} \) and \( \ln x \). This method is essentially the inverse of the product rule for differentiation.

In applying integration by parts to \( \int_{0}^{e} x^{p} \ln x \, dx \), we choose \( u = \ln x \) and \( dv = x^{p} \, dx \). By differentiating \( u \) and integrating \( dv \), the transformations are:

• \( du = \frac{1}{x} dx \)
• \( v = \frac{x^{p+1}}{p+1} \)

The integration by parts formula \(\int u \, dv = uv - \int v \, du\) then simplifies our original integral, tackling it through smaller, more manageable parts.

Once applied correctly, integration by parts reveals further insights into the integral's behavior and confirms convergence when \( p > 0 \), as evident by substituting these values back.

Behavior of Logarithmic Functions

Understanding logarithmic functions is crucial when dealing with integrals of the form \( x^{p} \ln x \). Here, we emphasize the behavior of \( \ln x \) near the limits of integration. At the lower limit of integration, \( x \) approaches zero. In this scenario, the logarithmic function \( \ln x \) heads to negative infinity. This tends to suggest divergence unless the accompanying function, \( x^{p} \), acts to limit this tendency.

For logarithmic functions,

• \( \ln x \to -\infty \) as \( x \to 0 \)
• \( \ln x = 1 \) at \( x = e \)

Understanding this behavior helps us establish that while \( \ln x \) can contribute to divergence at the lower limit, the positive power \( p > 0 \) in \( x^{p} \) ensures convergence by forcing the integrand towards zero faster than \( \ln x \) increases.

Thus, recognizing these dynamical changes in logarithmic behavior equips us to assess convergence more robustly.

Visualization of Integrals

Visualizing integrals helps us gain intuitive insights into how they behave across different parameter values. For the integral \( \int_{0}^{e} x^{p} \ln x \, dx \), plotting the function \( x^{p} \ln x \) for several \( p \) values clarifies convergence concerns.

Trying different values of \( p \), such as \( p = 1/2, 0, -1/2 \), can make the effects clear. For \( p = 1/2 \), the plot reveals a gentle curve that tends towards zero near the lower limit, indicating convergence. At \( p = 0 \), the plot may show marginal convergence, while negative \( p \) values, such as \( p = -1/2 \), clearly illustrate divergence as the function shoots towards negative infinity.

Through these visual explorations:

• We identify the influence of \( p \) on the integral's behavior.
• We confirm analytical results regarding convergence.

Visualization thus substantiates our computational findings and enriches our understanding.