Show limit of improper integral is 0

In summary: It tells you that if you have a function that is decreasing in one variable and increasing in another, then there is a point where the function is exactly equal for all values of the increasing variable.
  • #1
tjkubo
42
0

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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  • #2
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.
 
  • #3
I don't know how that would help since the set we're integrating over is changing with the limit.
 
  • #4
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?
 

What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite or the function being integrated has a discontinuity within the interval of integration.

Why is it important to show that the limit of an improper integral is 0?

Showing that the limit of an improper integral is 0 is important because it indicates that the function being integrated approaches 0 as the limits of integration approach infinity or a discontinuity. This ensures that the integral is convergent and can be evaluated accurately.

How do you show that the limit of an improper integral is 0?

To show that the limit of an improper integral is 0, you need to evaluate the integral using a limit as the upper or lower bound approaches infinity or the point of discontinuity. If the resulting limit is 0, then the limit of the improper integral is also 0.

What happens if the limit of an improper integral is not 0?

If the limit of an improper integral is not 0, then the integral is considered divergent and cannot be evaluated accurately. In this case, other methods such as using a different set of limits or using a different method of integration may need to be employed to accurately evaluate the integral.

Can the limit of an improper integral be a finite number other than 0?

Yes, the limit of an improper integral can be a finite number other than 0. This would indicate that the function being integrated approaches that finite number as the limits of integration approach infinity or a point of discontinuity. In this case, the integral is still considered convergent and can be evaluated accurately.

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