Problem 31 - A Course in Calculus and Real Analysis

Question

Show that \(\int_{1}^{\infty}\left(\cos t / t^{p}\right) d t\) and \(\int_{1}^{\infty}\left(\sin t / t^{p}\right) d t\) are absolutely convergent if \(p>1\) and that they are conditionally convergent if \(0

Step-by-Step Solution

Answer

The given integrals are absolutely convergent if \(p > 1\) because the integrals of the absolute value of the functions can be compared with the convergent integral \(\frac{1}{t^p}\) for \(p > 1\). They are conditionally convergent if \(0

Step 1: Absolute convergence

First, let's analyze the absolute convergence of both integrals. 1. Integral of absolute value of \(\frac{\cos(t)}{t^p}\): Given: \(\int_{1}^{\infty}\left|\frac{\cos(t)}{t^p}\right|dt\) Since \(|\cos(t)|\leq 1\), we can compare the given integral with the integral of \(\frac{1}{t^p}\): \(\int_{1}^{\infty}\frac{1}{t^p}dt=\frac{1}{p-1}\) for \(p > 1\). Thus, if \(p > 1\), the integral of the absolute value is convergent. 2. Integral of absolute value of \(\frac{\sin(t)}{t^p}\): Given: \(\int_{1}^{\infty}\left|\frac{\sin(t)}{t^p}\right|dt\) As \(|\sin(t)|\leq 1\), we can compare the given integral with the same integral \(\frac{1}{t^p}\): \(\int_{1}^{\infty}\frac{1}{t^p}dt=\frac{1}{p-1}\) for \(p > 1\). Thus, if \(p > 1\), the integral of the absolute value is convergent. Now, we've shown that both integrals are absolutely convergent if \(p>1\).

Step 2: Conditional convergence

Now, let's analyze the conditional convergence of both integrals. 1. Integral of \(\frac{\cos(t)}{t^p}\): To show the conditional convergence of this integral, we need to show that it is convergent while the integral of the absolute value is not convergent. We can use the Limit Comparison Test and the Analysis above for this: \(\int_{1}^{\infty}\left| \frac{\cos(t)}{t^p}\right| dt\) is not convergent for \(0

Key Concepts

Absolute Convergence

Understanding absolute convergence is crucial when dealing with improper integrals. An integral is said to be absolutely convergent if the integral of the absolute value of the function is convergent. For the given functions \( \cos(t)/t^p \) and \( \sin(t)/t^p \) in the exercise, the integrals of their absolute values compare with \( 1/t^p \) for \( p > 1 \).

Since both cosine and sine functions are bounded by 1, the absolute value of the integrals does not exceed the integral of \( 1/t^p \), which converges if \( p > 1 \) as it becomes a p-series with \( p > 1 \) that is known for its convergence. This concept allows students to easily determine the absolute convergence of many improper integrals. The comparison with a p-series is a powerful tool, simplifying the otherwise complex process of evaluating improper integrals.

Conditional Convergence

In contrast to absolute convergence, conditional convergence exists when the integral itself is convergent, but the integral of its absolute value is not. For the exercise provided, the integrals of \( \cos(t)/t^p \) and \( \sin(t)/t^p \) within the domain \( 0
Here, we are in a tricky spot; while the functions don't diverge wildly because of the bounded nature of cosine and sine, the p-series comparison shows divergence when \( p\leq1 \). Conditional convergence is subtle and requires a delicate analysis which can be performed using tools like Dirichlet's Test which we will discuss in the next section. It is important for students to recognize the distinct difference between absolute and conditional convergence as it impacts the behavior and the outcome of the improper integrals substantially.

Dirichlet's Test

Applying Dirichlet's Test

Dirichlet's Test is vital when handling challenging convergence tests for integrals. It states that for a certain range where the integral of the absolute value diverges, the integral of the function itself may still converge, given specific criteria are met. One of these criteria is that the integral of a function, say \( g(t) \), must be bounded if another function, like \( f(t)\), tends towards zero.

For both \( \cos(t)/t^p \) and \( \sin(t)/t^p \) when \( 0

Improper Integral Convergence

The concept of improper integral convergence is crucial to understand for various functions that are not limited to standard boundaries. An integral is referred to as 'improper' if either the interval of integration is infinite, or the function has an infinite discontinuity in the interval. In the textbook example, we see improper integrals with infinite intervals as we integrate from 1 to infinity.

Identifying convergence in these cases often requires comparing the function to a known convergent or divergent integral, testing absolute values, or using convergence tests like Dirichlet's. Without these methods, determining the behavior of these improper integrals would be much more complex. By framing the problem concisely and using the correct tests, students can break down the process and understand the behavior of integrals over infinite intervals or those with discontinuities.

Previous

Problem 29

Next

Problem 32

Other exercises in this chapter

Problem 28

Let \(f, g:[2, \infty) \rightarrow \mathbb{R}\) be defined by $$ f(t):=\left\\{\begin{array}{ll} 1 & \text { if } k \leq t

View solution

Problem 29

Let \(a \in \mathbb{R}\) and \(f:[a, \infty) \rightarrow \mathbb{R}\) be such that \(f\) is integrable on \([a, x]\) for all \(x \geq a\). Prove the following:

View solution

Problem 32

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:

View solution

Problem 33

Let \(g:[1, \infty) \rightarrow \mathbb{R}\) be a continuous real-valued function such that the function \(G:[a, \infty) \rightarrow \mathbb{R}\) defined by \(G

View solution

Key Concepts

Absolute Convergence

Understanding absolute convergence is crucial when dealing with improper integrals. An integral is said to be absolutely convergent if the integral of the absolute value of the function is convergent. For the given functions \( \cos(t)/t^p \) and \( \sin(t)/t^p \) in the exercise, the integrals of their absolute values compare with \( 1/t^p \) for \( p > 1 \).

Since both cosine and sine functions are bounded by 1, the absolute value of the integrals does not exceed the integral of \( 1/t^p \), which converges if \( p > 1 \) as it becomes a p-series with \( p > 1 \) that is known for its convergence. This concept allows students to easily determine the absolute convergence of many improper integrals. The comparison with a p-series is a powerful tool, simplifying the otherwise complex process of evaluating improper integrals.

Conditional Convergence

In contrast to absolute convergence, conditional convergence exists when the integral itself is convergent, but the integral of its absolute value is not. For the exercise provided, the integrals of \( \cos(t)/t^p \) and \( \sin(t)/t^p \) within the domain \( 0
Here, we are in a tricky spot; while the functions don't diverge wildly because of the bounded nature of cosine and sine, the p-series comparison shows divergence when \( p\leq1 \). Conditional convergence is subtle and requires a delicate analysis which can be performed using tools like Dirichlet's Test which we will discuss in the next section. It is important for students to recognize the distinct difference between absolute and conditional convergence as it impacts the behavior and the outcome of the improper integrals substantially.

Dirichlet's Test

Applying Dirichlet's Test

Dirichlet's Test is vital when handling challenging convergence tests for integrals. It states that for a certain range where the integral of the absolute value diverges, the integral of the function itself may still converge, given specific criteria are met. One of these criteria is that the integral of a function, say \( g(t) \), must be bounded if another function, like \( f(t)\), tends towards zero.

For both \( \cos(t)/t^p \) and \( \sin(t)/t^p \) when \( 0

Improper Integral Convergence

The concept of improper integral convergence is crucial to understand for various functions that are not limited to standard boundaries. An integral is referred to as 'improper' if either the interval of integration is infinite, or the function has an infinite discontinuity in the interval. In the textbook example, we see improper integrals with infinite intervals as we integrate from 1 to infinity.

Identifying convergence in these cases often requires comparing the function to a known convergent or divergent integral, testing absolute values, or using convergence tests like Dirichlet's. Without these methods, determining the behavior of these improper integrals would be much more complex. By framing the problem concisely and using the correct tests, students can break down the process and understand the behavior of integrals over infinite intervals or those with discontinuities.

Previous

Problem 29

Next

Problem 32

Other exercises in this chapter

Problem 28

Let \(f, g:[2, \infty) \rightarrow \mathbb{R}\) be defined by $$ f(t):=\left\\{\begin{array}{ll} 1 & \text { if } k \leq t

View solution

Problem 29

Let \(a \in \mathbb{R}\) and \(f:[a, \infty) \rightarrow \mathbb{R}\) be such that \(f\) is integrable on \([a, x]\) for all \(x \geq a\). Prove the following:

View solution

Problem 32

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:

View solution

Problem 33

Let \(g:[1, \infty) \rightarrow \mathbb{R}\) be a continuous real-valued function such that the function \(G:[a, \infty) \rightarrow \mathbb{R}\) defined by \(G

View solution