Question

If \(\lim _{x \rightarrow 0^{+}} f(x)=\infty\), we define $$ \int_{0}^{\infty} f(x) d x=\lim _{c \rightarrow 0^{-}} \int_{c}^{1} f(x) d x+\lim _{b \rightarrow \infty} \int_{1}^{b} f(x) d x $$ provided both limits exist. Otherwise, we say that \(\int_{0}^{\infty} f(x) d x\) diverges. Show that \(\int_{0}^{\infty} \frac{1}{x^{p}} d x\) diverges for all \(p\).

Short answer

The integral diverges for all \(p\).

Step-by-step solution

Write the Given Integral

We are given the improper integral \[ \int_{0}^{\infty} \frac{1}{x^p} \, dx. \] We need to evaluate this integral based on the limits provided in the definition.

Evaluate the Integral from c to 1

First, consider \[ \lim_{c \to 0^+} \int_{c}^{1} \frac{1}{x^p} \, dx. \] Integrating, we have:\[ \int \frac{1}{x^p} \, dx = \begin{cases}\frac{x^{1-p}}{1-p}, & \text{if } p eq 1, \\ln|x|, & \text{if } p = 1.\end{cases} \]Thus:\[ \int_{c}^{1} \frac{1}{x^p} \, dx = \begin{cases}\left[\frac{x^{1-p}}{1-p}\right]_{c}^{1}, & p eq 1, \\left[\ln|x|\right]_{c}^{1}, & p = 1.\end{cases} \]

Analyze the Limit as c Approaches 0 from the Right

For \(p eq 1\):\[ \lim_{c \to 0^+} \left(\frac{1}{1-p} - \frac{c^{1-p}}{1-p}\right). \]If \(p < 1\), the term \(c^{1-p}\) tends to infinity as \(c\) approaches 0. If \(p > 1\), the term \(c^{1-p}\) stays finite and tends to 0 as \(c\) approaches 0.For \(p = 1\):\[ \lim_{c \to 0^+} \left(0 - \ln(c)\right), \] which tends to infinity as \(c\) approaches 0.

Evaluate the Integral from 1 to b

Now, consider \[ \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^p} \, dx. \]Using our integral result from Step 2: \[ \int_{1}^{b} \frac{1}{x^p} \, dx = \begin{cases}\left[\frac{x^{1-p}}{1-p}\right]_{1}^{b}, & p eq 1, \\left[\ln|x|\right]_{1}^{b}, & p = 1.\end{cases} \]So:\[ \frac{b^{1-p}}{1-p} - \frac{1^{1-p}}{1-p}, \] or \[ \ln(b) - \ln(1) \] for \(p = 1\).

Analyze the Limit as b Approaches Infinity

For \(p eq 1\):\[ \lim_{b \to \infty} \left(\frac{b^{1-p}}{1-p} - \frac{1}{1-p}\right). \]If \(p < 1\), this limit diverges to infinity because \(b^{1-p}\) grows without bound. If \(p > 1\), the limit converges to a finite value since \(b^{1-p}\) tends to 0.For \(p = 1\):\[ \lim_{b \to \infty} \ln(b), \] which diverges to infinity as \(b\) approaches infinity.

Conclude on the Convergence of the Entire Integral

From the analysis, if \(p < 1\), the limit as \(c \to 0^+\) diverges, and if \(p \geq 1\), the limit as \(b \to \infty\) diverges. Therefore, for all \(p\), either the integral from \(c\) to \(1\) or from \(1\) to \(b\) diverges, leading \(\int_{0}^{\infty} \frac{1}{x^p} \, dx\) to diverge for all \(p\).

Key concepts

Divergence

Divergence in the context of improper integrals refers to the behavior of the integral when it does not converge to a finite number. For an integral to converge, it must have a limit as the bounds approach the point of infinity or some point of discontinuity. If no such limit exists, the integral is said to diverge.
A common scenario is when you deal with functions like \( \frac{1}{x^p} \) as \( x \to \infty \), especially when \( p \) varies.
Determining divergence requires evaluating the behavior of the integral's limit:

• If it approaches infinity, we say the integral diverges.
• If it settles at a finite value, it converges.

Understanding divergence is crucial in defining whether an improper integral has a finite value or not.

Limit Evaluation

Evaluating limits is essential in the study of improper integrals, particularly because it helps determine the convergence or divergence of the integral. The limits help us understand the behavior of the integral as the variable approaches infinity or a point of discontinuity.
In the case of the integral \( \int_{0}^{\infty} \frac{1}{x^p} \, dx \), we consider two limits:

• \( \lim_{c \to 0^+} \int_{c}^{1} \frac{1}{x^p} \, dx \)
• \( \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^p} \, dx \)

The key is in evaluating these limits for different values of \( p \):
- If \( p = 1 \), the integral involves \( \ln(x) \), and it diverges as \( x \to 0 \) or \( x \to \infty \).
- If \( p < 1 \), the function becomes unbounded as \( c \to 0 \).
- If \( p > 1 \), the limit as \( b \to \infty \) involves a term that approaches zero, which indicates potential convergence, but the other limit is divergent.
Understanding how to properly evaluate these limits is fundamental for concluding the convergence or divergence of an integral.

Improper Integrals with Parameter p

When evaluating improper integrals with a parameter \( p \), we must understand how the value of \( p \) affects the result. The classic example \( \int_{0}^{\infty} \frac{1}{x^p} \, dx \) demonstrates how \( p \) influences whether the integral diverges or converges.
Improper integrals are divided based on the range of \( p \):

• For \( p = 1 \), the integral involves logarithms, specifically \( \ln(x) \), which leads to divergence.
• For \( p < 1 \), the integral diverges because the term tends to infinity as \( x \to 0 \).
• For \( p > 1 \), although one part converges, the integral tends to zero for large values of \( x \), causing divergent behavior as a whole due to the other side of the evaluation.

This exploration shows that for any \( p \), the integral will diverge due to at least one problematic limit. The parameter \( p \) affects the power of \( x \) and thus influences the convergence behavior significantly.