Problem 32 - A Course in Calculus and Real Analysis

Question

Let \(f:[1, \infty) \rightarrow \mathbb{R}\) be such that \(f\) is integrable on \([1, x]\) for every \(x \geq 1\). Prove the following using Proposition 9.42: (i) If there are \(p>1\) and \(\ell \in \mathbb{R}\) such that \(t^{p} f(t) \rightarrow \ell\) as \(t \rightarrow \infty\), then \(\int_{1}^{\infty} f(t) d t\) is absolutely convergent. (ii) Suppose \(f(t)>0\) for all \(t \in[1, \infty)\). If there are \(p \leq 1\) and \(\ell \neq 0\) such that \(t^{p} f(t) \rightarrow \ell\) as \(t \rightarrow \infty\), then \(\int_{1}^{\infty} f(t) d t\) is divergent.

Step-by-Step Solution

Answer

Proof for (i): As \(t \rightarrow \infty\), we have \(\left| t^{p} f(t) - \ell \right| < 1\). From this inequality, we can deduce that \(\left| f(t) \right| < \frac{\ell + 1}{t^{p}}\). By the Comparison Test, since \(p>1\), the integral \(\int_{1}^{\infty} \frac{\ell + 1}{t^{p}} \, dt\) is convergent, implying that \(\int_{1}^{\infty} \left| f(t) \right| \, dt\) is convergent as well. Therefore, \(\int_{1}^{\infty} f(t) \, dt\) is absolutely convergent. Proof for (ii): As \(t \rightarrow \infty\), \(t^{p} f(t) \rightarrow \ell \neq 0\), which implies that \(f(t)\) is asymptotically equal to \(\frac{\ell}{t^{p}}\). Since \(p \leq 1\), the integral \(\int_{1}^{\infty} \frac{\ell}{t^{p}} \, dt\) is divergent. Hence, \( \int_{1}^{\infty} f(t) \, dt \) is also divergent as \(f(t) > 0\) and is asymptotically equal to \(\frac{\ell}{t^{p}}\).

Step 1: Rewrite the given condition

Given that as \( t \rightarrow \infty \), we have \( t^{p} f(t) \rightarrow \ell \), implying that for sufficiently large \( t \), \(\left| t^{p} f(t) - \ell \right| < 1 \). This can be rewritten as \(-1 < t^{p} f(t) - \ell < 1\)

Step 2: Re-arrange the inequality

We can then isolate \( t^{p}f(t) \) which yields \(\ell - 1 < t^{p}f(t) < \ell + 1\)

Step 3: Divide through by \( t^{p} \) and use absolute values

Dividing through by \( t^{p} \), we get \(\frac{\ell - 1}{t^{p}} < f(t) < \frac{\ell + 1}{t^{p}}\) Using absolute values, we can write this as \(\left| f(t) \right| < \frac{\ell + 1}{t^{p}}\)

Step 4: Use Comparison Test

By the Comparison test, because \( p>1 \), the integral \( \int_{1}^{\infty} \frac{\ell + 1}{t^{p}} \, dt = \ell + 1 \times \int_{1}^{\infty} t^{-p} \, dt \) is convergent. Since \( \left| f(t) \right| \) is less than this convergent integral, it follows that \( \int_{1}^{\infty} \left| f(t) \right| \, dt \) is convergent as well. That is, \( \int_{1}^{\infty} f(t) \, dt \) is absolutely convergent. Proof for (ii)

Step 1: Rewrite the given condition

Given that as \( t \rightarrow \infty \), we have \( t^{p} f(t) \rightarrow \ell \neq 0 \), means that for sufficiently large \( t \), \( t^{p} f(t) \) approaches a non-zero constant.

Step 2: Find an expression for \( f(t) \)

From the limit condition, we can deduce that \( f(t) \) is asymptotically equal to \( \frac{\ell}{t^{p}} \).

Step 3: Use Comparison Test

By the Comparison Test, since \( p \leq 1 \), the integral \(\int_{1}^{\infty} \frac{\ell}{t^{p}} \, dt = \ell \times \int_{1}^{\infty} t^{-p} \, dt \) is divergent. Hence, since \( f(t) > 0 \) and is asymptotically equal to \( \frac{\ell}{t^{p}} \), \( \int_{1}^{\infty} f(t) \, dt \) is divergent too.

Key Concepts

Absolute Convergence

Understanding the concept of absolute convergence is crucial when dealing with series and integrals. In real analysis, a function or series is said to absolutely converge if the sum or integral of the absolute values is finite.

This concept is significant because if a series or an integral is absolutely convergent, then it is also convergent. However, the converse is not always true; there are series and integrals that converge but do not converge absolutely, which can lead to different properties and behaviors.

For example, in the exercise, if a function f(t) that is integrable on [1, x] satisfies a condition that, as t approaches infinity, t^p f(t) approaches a limit ℓ for some p > 1, this suggests that f(t) decreases sufficiently fast to ensure the convergence of the integral from 1 to infinity of f(t). Hence, the integral is said to be absolutely convergent.

Comparison Test

The comparison test is a fundamental tool in determining the convergence or divergence of an improper integral or series. It works by comparing the terms of a sequence or the integrand of an integral to another sequence or integrand whose convergence property is known.

For improper integrals, if there exists a function g(t) that is always greater than |f(t)| such that the integral of g(t) is known to be convergent, then the integral of |f(t)| is also convergent. Conversely, if g(t) is always less than f(t) and its integral is divergent, then the integral of f(t) is also divergent.

In the given exercise solution, the comparison test is applied to show that the integral of |f(t)| is less than the integral of a function (ℓ + 1)/t^p known to be convergent, and this yields the absolute convergence of the original integral.

Improper Integrals

Improper integrals arise when the integrand function or the limits of integration involve infinity, or when the integrand has an infinite discontinuity within the limits of integration. To evaluate an improper integral, one typically converts it to a limit of proper integrals.

These kinds of integrals are incredibly important in advanced calculus, physics, and engineering, as they often appear in real-world scenarios. In the context of the presented exercise, the improper integral of f(t) from 1 to infinity needs to be evaluated, and due to its nature, one must assess whether this integral is convergent or not. This is often done using techniques such as the comparison test, as explained previously.

Convergence and Divergence

Convergence and divergence are the fundamental concepts of an integral or series that describe their behavior as they extend to infinity or as their terms grow without bounds. A sequence, series, or integral converges if it approaches a particular value as its terms increase or the limits of integration go to infinity. If it does not approach a specific value, it diverges.

These properties are especially relevant for improper integrals where the limits of integration are unbounded. In such cases, determining whether an integral converges or diverges involves finding whether the sum of its values stabilizes to a finite number or increases indefinitely.

For the cases discussed in the problem at hand, the key lies in understanding how to evaluate the behavior of f(t) and applying appropriate tests like the comparison test to establish the convergence or divergence of the integrals in question.