# Does this improper integral converge?

1. Feb 5, 2005

### JoePharos

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?

2. Feb 5, 2005

### dextercioby

Take for example $g(x)=\frac{1}{x^{2}}$ and see what happens...

Daniel.

3. Feb 5, 2005

### JoePharos

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. Feb 6, 2005

### TimNguyen

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.