Date: 9/14/95 at 7:15:15 From: Filiberto Strazzari Subject: Generalized Integrals Let's consider the function 1 f(x) = ----------------- x^h * (log x)^k For what real values of h and k does infinite 1 / / | f(x) dx and | f(x) dx / / 1 0 converge or diverge? Actually, I should add I'm in little hurry about this problem. Thank you very much for your effort. Filiberto Strazzari <firstname.lastname@example.org>
Date: 10/11/95 at 17:17:39 From: Doctor Ethan Subject: Re: Generalized Integrals Well, here are my thoughts on your problem. One way to phrase this question is "For what values of h and k does the bottom grow fast enough so that the integral converges?" Let's look at integral one. Both things on the bottom are positive for all of these values, so there is no fear that they will cancel each other out. When h>1 1/x^h will converge by itself, so adding that extra stuff for any k will be okay. So look at h<=1. You may know that 1/x does not converge, so one way to think about this is, "When is log[x]^k going to grow faster than x in the long run? " We will show that the answer is that it never will. The notion of "grow faster" is very closely related to the derivative, so let us compare the derivative of the two functions. The derivative of x is 1 and the derivative of log[x]^k is k * log[x]^(k-1) *1/x . But as x approaches infinity this approaches zero. (You can prove this using L'Hopital's rule.) So this tells that for the first one, when h<=1 this will not converge for any value of k. Great! Now, on to the second one. I will just give you a hint and let you see if you can figure it out. We know that the polynomial part (i.e. x^h) will have finite value for h=0 or greater. So for those you can just examine possible k values in a similar way to what we did earlier. Now think about the values for h<0. These start to get interesting. Think about them. Good Luck... - Doctor Ethan, The Geometry Forum
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