Question

Determine whether the following statements are true and give an explanation or counterexample. a. If \(f\) is continuous and \(0p\). d. If \(\int_{1}^{\infty} x^{-p} d x\) exists, then \(\int_{1}^{\infty} x^{-q} d x\) exists, where \(q>p\). e. \(\int_{1}^{\infty} \frac{d x}{x^{3 p+2}}\) exists, for \(p>-\frac{1}{3}\).

Short answer

#Short Answer# a. True. The comparison test shows that if \(0 < f(x) < g(x)\) and \(\int_{0}^{\infty} g(x) dx\) converges, then \(\int_{0}^{\infty} f(x) dx\) also converges. b. False. The function f(x) = 1 provides a counterexample where the limit equals 1, but the integral does not exist. c. True. If \(\int_{0}^{1} x^{-p} dx\) exists (i.e., \(p < 1\)), then for \(q > p\), \(\int_{0}^{1} x^{-q} dx\) also exists. d. True. If \(\int_{1}^{\infty} x^{-p} dx\) exists (i.e., \(p > 1\)), then for \(q > p\), \(\int_{1}^{\infty} x^{-q} dx\) also exists. e. True. The integral \(\int_{1}^{\infty} \frac{dx}{x^{3p+2}}\) exists if \(p > -\frac{1}{3}\).

Step-by-step solution

a. If \(f\) is continuous and \(0

True. Because f(x) is continuous and positive, and its bounds are within the bounds of g(x), we can apply the comparison test. Since the integral of g(x) converges to M, which is finite, it follows that the integral of f(x) also converges.

b. If \(\lim _{x \rightarrow \infty} f(x)=1,\) then \(\int_{0}^{\infty} f(x) d x\) exists.

False. Consider the counterexample f(x) = 1. Here, \(\lim_{x\rightarrow\infty} f(x) = 1\), but the integral \(\int_0^{\infty} f(x)\,dx\) equals infinity, so it does not exist.

c. If \(\int_{0}^{1} x^{-p} d x\) exists, then \(\int_{0}^{1} x^{-q} d x\) exists, where \(q>p\).

True. If \(\int_{0}^{1} x^{-p} dx\) exists, this means \(p < 1\) (since the integral converges for \(p < 1\) and diverges for \(p \geq 1\)). Now, if \(q > p\) and \(p < 1\), then definitely \(q > 0\). For any \(0 < q < 1\), the integral \(\int_{0}^{1} x^{-q} dx\) will also exist. Therefore, if the integral with an exponent -p exists, then the integral with an exponent -q also exists, where q > p.

d. If \(\int_{1}^{\infty} x^{-p} d x\) exists, then \(\int_{1}^{\infty} x^{-q} d x\) exists, where \(q>p\).

True. If \(\int_{1}^{\infty} x^{-p} dx\) exists, this means \(p > 1\) (since the integral converges for \(p > 1\) and diverges for \(p \leq 1\)). Now, if \(q > p\) and \(p > 1\), then definitely \(q > 1\). For any \(q > 1\), the integral \(\int_{1}^{\infty} x^{-q} dx\) will also exist. Therefore, if the integral with an exponent -p exists, then the integral with an exponent -q also exists, where q > p.

e. \(\int_{1}^{\infty} \frac{d x}{x^{3 p+2}}\) exists, for \(p>-\frac{1}{3}\).

True. Notice that the integral can be written as \(\int_{1}^{\infty}x^{-(3p+2)}\,dx\). For the integral to exist, we need the exponent of x to be greater than 1. This can be written as the inequality \((-3p + 2) > 1\), which simplifies to \(p > -\frac{1}{3}\). Therefore, if \(p > -\frac{1}{3}\), the integral exists.

Key concepts

Improper Integrals

Improper integrals are a type of integral that extend over an unbounded region or approach a point where the function becomes undefined. They are typically represented as an integral over an infinite interval, such as \(\int_{a}^{\infty} f(x) \,dx\), or as an integral where the function has a singularity (a point where it 'blows up' or becomes undefined) within the interval, like \(\int_{a}^{b} g(x) \,dx\) where g(x) is undefined at \(x=b\). Because improper integrals often represent infinite sums or quantities, evaluating their convergence—whether they result in a finite number—is a critical topic in calculus.

Improper integrals are essential in various fields including physics, probability, and engineering, as they allow for the calculation of quantities that are otherwise difficult to define using traditional methods. For example, they can represent the total mass of an infinite wire with a given density function, or the probability of an event occurring in a continuous probability distribution.

Comparison Test for Integrals

The comparison test for integrals is a useful method for determining the convergence or divergence of improper integrals. When we have an integral of a non-negative function \(f\), and it's difficult to evaluate directly, we can instead compare it to another function \(g\) that is easier to integrate and for which we know the behavior of its integral.

The basic idea is that if \(0 \leq f(x) \leq g(x)\) for all \(x\) in the interval of integration, and if \(\int g(x)\,dx\) converges, then \(\int f(x)\,dx\) also converges. Conversely, if \(\int f(x)\,dx\) diverges, then \(\int g(x)\,dx\) must also diverge. This test is handy when dealing with non-standard functions or functions without elementary antiderivatives.

Convergence and Divergence of Series

The convergence or divergence of a series is a fundamental concept that helps determine whether an infinite sum has a finite limit (converges) or not (diverges). A series is an infinite sum of the terms of a sequence, given by \(\sum_{n=1}^{\infty} a_n\). Whether a series converges or diverges can depend on various factors, such as the rate at which the terms \(a_n\) approach zero or grow without bound.

Various tests, including comparison tests, ratio tests, and root tests, can help determine a series's behavior. For instance, the famous harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\) diverges, while the geometric series \(\sum_{n=0}^{\infty} ar^n\) for \(|r| < 1\) converges. Understanding the convergence and divergence of series is crucial not only in pure mathematics but also in applications like signal processing and quantum physics.

Integral Convergence Criteria

Determining the convergence of an integral involves verifying whether the integral yields a finite value. Integral convergence criteria are rules or tests used to assess whether a given integral will converge. One such criterion is the behavior of the function as it approaches an infinite limit or a point of discontinuity. For instance, if the integral \(\int_{1}^{\infty} f(x)\,dx\) is considered, and \(f(x)\) decreases and approaches zero as \(x\) approaches infinity, it is more likely that the integral will converge.

Another criterion is the p-test, which states that the integral \(\int_{1}^{\infty} x^{-p}\,dx\) converges if \(p > 1\) and diverges if \(p \leq 1\). This test can be particularly useful for integrals involving power functions. The integral convergence criteria equip students with the tools to analyze and engage with a variety of challenging integrals, offering a deeper insight into the underlying mathematics and its practical applications.