Recent content by 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...

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...