# Homework Help: If integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) converge

1. Feb 14, 2010

### no_alone

1. The problem statement, all variables and given/known data
f(x) positive and continuous in [a,$$\infty$$)
prove of disprove:
if $$\int_a^\infty f(x)dx$$ converge there is a 0<c<1 so that
$$\int_a^\infty f(x)^p dx$$ converge for every c$$\leq$$p$$\leq$$1

2. Relevant equations
every thing in calculus 1+2 no two vars ..

3. The attempt at a solution
Well I think that there is ..
But i have no clue on how to solve it....
Thank you.

2. Feb 14, 2010

### Dick

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

I don't think it's true. Try find an example of an f(x) that just barely converges and f(x)^p diverges for all 0<p<1.

3. Feb 15, 2010

### no_alone

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

Ok
$$\frac{1}{xln^2(x)}$$
for every p<1
$$\frac{1}{x^pln^{2/p}(x)} > \frac{1}{x}$$

But i always have problem proving that for every p an z lim(x^p) > lim(ln^z(x))
Can you help me with this?

4. Feb 15, 2010

### rsa58

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

notice that you function is undefined at x=0 and the inequality you propose is incorrect for 0<x<1

5. Feb 15, 2010

### rsa58

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

sorry it is completely incorrect

6. Feb 15, 2010

### rsa58

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

i have a feeling it is true since f(x)^p depends continously on p. are you sure this is all the info?

7. Feb 15, 2010

### no_alone

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

yes but I dont care about when 0<x<1 because its in a half closed section when f(X) is defined and continuous ,
I only look at infinity, and in infinity ,

for every p<1

$$\frac{1}{x^pln^{2/p}(x)} > \frac{1}{x}$$

8. Feb 15, 2010

### rsa58

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

incorrect try x=3. the inequality is actually backwards. plus your function cannot include p. the result must hold for ALL p. since f is positive you can use the ration

9. Feb 15, 2010

### Dick

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

That's a great choice! It would be enough to show that in the limit ln(x)<x^p for all p>0, wouldn't it? Just use l'Hopital on the limit of x^p/ln(x).

Last edited: Feb 15, 2010
10. Feb 16, 2010

### no_alone

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

Thank you very much dick..
It's not really homework question,
Its a example test question .
This question really on my mind because it was really against my intuition ,
And my usually my intuition is true ,(my problem is proving it)
Can you supply another f(X) that is good?,
Thank you.

11. Feb 16, 2010

### Dick

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

You're welcome. 'Intuition' can go a little wrong when you are dealing with an 'infinite' part in the problem. Guess I don't have a ready stock of problems. You might scan through other forum posts looking for interesting problems. There are some in there. And feel free to help the poster if you've got a good idea on how to solve it.

12. Feb 16, 2010

### no_alone

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

Sorry dick but I think you misunderstood me,
I asked if you can give me another counter example beside this one I supplied?
Thank you.

13. Feb 16, 2010

### Dick

Re: if integral f(x) converge there is a c<1 that for all c<p<1 integral f^p(x) conve

Nothing pops into my head. There's 1/(log(x)^n*x) for n>1. But that's just a variation on the same theme.