Improper integral convergence and implications of infinite limits

In summary, the conversation discusses the proof that if a continuous function on [1,∞) has a limit of α at infinity and its integral from 1 to ∞ converges, then α must be equal to 0. The discussion includes the definition of an improper integral and the attempt at a solution using the epsilon-delta version of a limit. However, the approach taken may not be correct and the use of N instead of δ is suggested.
  • #1
Szichedelic
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0

Homework Statement



Let f be a continuous function on [1,∞) such that [itex]\lim_{x\rightarrow ∞}[/itex]f(x)=α. Show that if the integral [itex]\int^{∞}_{1} f(x)dx[/itex] converges, then α must be 0.

Homework Equations


Definition of an Improper Integral
Let f be a continuous function on an interval [a,∞). then we define [itex]\int^{∞}_{a} f(x)dx[/itex] to be [itex]\lim_{b \rightarrow ∞}\int^{b}_{a} f(x)dx[/itex] if this limit exists. In this case, the integral is said to be convergent.

The Attempt at a Solution



My attempt at this solution was first to start with the epsilon-delta version of a limit of a function. That is, since [itex]\lim_{x\rightarrow ∞}[/itex]f(x)=α, then [itex]\forall \epsilon[/itex]>0 with [itex]x_{0}\in [/itex] [1,∞) [itex]\exists~ \delta[/itex]>0 s.t. |f(x) - α | whenever |x-[itex]x_{0}[/itex]|<[itex]\delta[/itex]. Then, I assumed that α was not equal to zero. From there I split up my inequality so that

-ε<f(x) - α< ε [itex]\Rightarrow[/itex] α - ε<f(x)< ε + α [itex]\Rightarrow[/itex] [itex]\int^{∞}_{1}[/itex](α - ε)dx < [itex]\int^{∞}_{1} f(x)dx[/itex] < [itex]\int^{∞}_{1} (α + ε)dx[/itex]

From here, I am getting stuck in my logic and I think that perhaps this is not the route I want to go. I would appreciate it if I could be offered any hints on where to go from here, or on whether or not I should start over. Thanks!
 
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  • #2
Hi Szichedelic! :smile:
Szichedelic said:
My attempt at this solution was first to start with the epsilon-delta version of a limit of a function. That is, since lim[itex]_{x\rightarrow∞}[/itex]f(x)=α, then [itex]\forall[/itex] [itex]\epsilon[/itex]>0 with x[itex]_{o}[/itex] [itex]\in[/itex] [1,∞) [itex]\exists[/itex] [itex]\delta[/itex]>0 s.t. |f(x) - α | whenever |x-x[itex]_{o}[/itex]|<[itex]\delta[/itex].

No, for ∞, your δ is a number "near ∞" …

we usually use N instead, and we look for an N such that x > N => |f(x) - α| < ε :wink:
 

What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite or where the integrand is unbounded at one or more points within the interval of integration. This means that the area under the curve cannot be calculated using a traditional Riemann sum, and the integral must be approached using a limit as the boundaries of the interval approach infinity.

How do you determine if an improper integral converges or diverges?

To determine if an improper integral converges or diverges, you can use the comparison test or the limit comparison test. If the integral can be shown to be larger or smaller than a known convergent or divergent integral, then the original integral will have the same property. Alternatively, you can evaluate the limit of the integral as the boundaries approach infinity. If this limit exists and is finite, then the integral converges; if the limit is infinite or does not exist, then the integral diverges.

What are the implications of an improper integral having an infinite limit?

An improper integral with an infinite limit means that the area under the curve is infinite. This can have significant implications in real-world scenarios, as it implies that there is an infinite amount of something being measured, such as infinite volume or infinite mass. It also means that the function being integrated is unbounded, which may indicate a problem with the model or equation being used to describe the situation.

Can an improper integral have a finite limit and still diverge?

Yes, an improper integral can have a finite limit and still diverge. This can occur when the function being integrated has a vertical asymptote within the interval of integration, causing the limit of the function as it approaches the asymptote to be infinite. In this case, the integral may have a finite limit, but the function itself is unbounded and therefore the integral diverges.

What are some real-world applications of improper integrals?

Improper integrals have many real-world applications, especially in physics and engineering. For example, they can be used to calculate the center of mass of an object with a varying density, the work done by a force that varies over distance, and the amount of fluid in a tank with a leak. They can also be used in probability and statistics to calculate the expected value of a continuous random variable.

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