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Homework Help Overview

The discussion revolves around the limit of an improper integral involving a real-valued, bounded, continuous, and non-negative function \( f \). Participants are exploring whether the limit of the integral \( \int_x^\infty f(t)\,dt \) approaches 0 as \( x \) approaches infinity, given that the integral is convergent for all \( x \).

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to prove the limit intuitively seems true and are considering the implications of the convergence of the integral. There are questions about how to establish that \( \lim_{t\rightarrow \infty} f(t) = 0 \) and the relationship between the limit and the existence of the improper integral. One participant suggests looking into the dominated convergence theorem.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts and questioning the applicability of the dominated convergence theorem in this context. There is no explicit consensus yet, but some guidance has been offered regarding the theorem's potential relevance.

Contextual Notes

Participants are navigating the complexities of the problem, including the changing limits of integration and the conditions under which the improper integral converges. There is uncertainty about how to formally prove certain aspects of the problem.

tjkubo
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Homework Statement


Suppose [itex]f[/itex] is real-valued, bounded, continuous, and non-negative and suppose [itex]\int_x^\infty f(t)\,dt[/itex] is convergent (is finite) for all [itex]x[/itex]. Is it true that
[tex]\lim_{x\rightarrow \infty} {\int_x^\infty f(t)\,dt} = 0 \ ?[/tex]

Homework Equations



The Attempt at a Solution


I can't think of a counterexample and it seems true intuitively, so I'm trying to prove it's true.
Given [itex]\epsilon >0[/itex], I want to show there is [itex]M[/itex] such that
[tex]\int_M^\infty f(t)\,dt < \epsilon \ .[/tex]
I think [itex]\lim_{t\rightarrow \infty} {f(t)} = 0[/itex] although I am not sure how to prove this. Specifically, how would you prove that the improper integral would not exist if this limit did not exist?
I'm not sure what to do at this point.
 
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tjkubo said:

Homework Statement


Suppose [itex]f[/itex] is real-valued, bounded, continuous, and non-negative and suppose [itex]\int_x^\infty f(t)\,dt[/itex] is convergent (is finite) for all [itex]x[/itex]. Is it true that
[tex]\lim_{x\rightarrow \infty} {\int_x^\infty f(t)\,dt} = 0 \ ?[/tex]

Homework Equations



The Attempt at a Solution


I can't think of a counterexample and it seems true intuitively, so I'm trying to prove it's true.
Given [itex]\epsilon >0[/itex], I want to show there is [itex]M[/itex] such that
[tex]\int_M^\infty f(t)\,dt < \epsilon \ .[/tex]
I think [itex]\lim_{t\rightarrow \infty} {f(t)} = 0[/itex] although I am not sure how to prove this. Specifically, how would you prove that the improper integral would not exist if this limit did not exist?
I'm not sure what to do at this point.

Try looking at the dominated convergence theorem.
 
I don't know how that would help since the set we're integrating over is changing with the limit.
 
Ah, so you can use that theorem??

Note that

[tex]\int_x^{+\infty}{f(t)dt}=\int_{-\infty}^{+\infty}{I_{[x,+\infty[}(t)f(t)dt}[/tex]

with

[tex]I_{[x,+\infty[}:\mathbb{R}\rightarrow \mathbb{R}:t\rightarrow \left\{\begin{array}{cc} 0 & \text{if}~t<x\\ 1 & \text{if}~t\geq x \end{array}\right.[/tex]

So what does the dominated convergence theorem tell you?
 

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