Does this improper integral converge?

In summary, the question is whether the given improper integral of g(x) from A to +infinity converges or if it is impossible to determine. The given criteria state that g(x) can be any non-negative function that is less than f(x)=1/x. While it is known that for functions in the form of 1/(x^p) where p>1, the integral will converge, this is not necessarily the case for g(x). Therefore, the convergence or divergence of the integral depends on the specific function chosen for g(x).
  • #1
JoePharos
2
0
Does this improper integral converge?

I have an interesting problem that has been bothering me.

Given:
f(x)=1/x
g(x)= (any continuous, non-negative function)
g(x)<f(x)
A = (a positive constant)

I want to know:
Does the integral of g(x) from A to +infinity converge? Or is it impossible to determine?
 
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  • #2
Take for example [itex] g(x)=\frac{1}{x^{2}} [/itex] and see what happens...

Daniel.
 
  • #3
g(x) does NOT need to be in form 1/(x^P)

I already know that for any function of the form 1/(x^p), where p>1, it will converge. Thus, I know 1/(x^2) converges.

HOWEVER, g(x) does NOT need to be in the form of 1/(x^p). g(x) can be ANY non-negative function such that g(x)<f(x). That's a completely different picture.
 
  • #4
Assumingly, you're asking whether or not a function like 1/(x+1) converges or something? With the criteria you mentioned, the function where f(x) > g(x) and a non-negative function could be either/or convergent/divergent.
 

1. What is an improper integral?

An improper integral is an integral that does not have both limits of integration defined or has at least one limit of integration at infinity.

2. How do you determine if an improper integral converges?

To determine if an improper integral converges, you must evaluate the limit of the integral as the limits of integration approach the defined boundaries. If the limit exists and is a finite number, then the integral converges. Otherwise, it diverges.

3. Can an improper integral converge even if it has an infinite limit of integration?

Yes, an improper integral can still converge even if it has an infinite limit of integration. This depends on how the function behaves at the boundary and if the limit of the integral exists.

4. What is the difference between a convergent and a divergent improper integral?

A convergent improper integral is one that has a finite limit when evaluated at the boundaries, while a divergent improper integral does not have a finite limit at the boundaries.

5. Are there any standard methods for evaluating improper integrals?

Yes, there are several methods for evaluating improper integrals, including the comparison test, limit comparison test, and the integration by parts method. However, the method used will depend on the specific integral and its limits of integration.

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