Improper Integral Convergence of Positive Continuous Function I ask for some help or hint how to deal with this question:
Suppose f(x) is continuous and positive function for all $$x\ge a$$
Prove or provide a counterexample:
If $$\int_{a}^\infty f(x)dx $$ converges 
than there are $$0<c<1$$ that $$\int_{a}^\infty f^p(x)dx$$
converges for every $$c\le\ p \le1$$
Thanks.    
 A: Counter-example:
$$f(x)=\frac{1}{x\cdot (\log x)^2},\quad x\ge 2.$$
A: There is already an answer posted, but let me take the opportunity to talk about how you could approach the problem.
Assuming the integral converges, $f(x)$ will be quite small most of the time. But when $y$ is small and $0<p<1$, $y^p$ is larger than $y$. It is in fact much larger, in the sense that $y^p/y\to\infty$ when $y\to0$. Since $f(x)$ will be very small most of the time, $f^p(x)$ will be much larger than $f(x)$ most of the time.
This train of thoughts leads to a conjecture: There ought to be a counterexample.
Next, since powers are involved, it seems reasonable to look for the counterexample among power functions $f(x)=1/x^a$, for $a>1$ (if $a\le1$, the integral diverges). Playing around with this idea, you will see that it fails, but it comes very close to succeeding; the more the closer to $1$ you pick $a$.
And that is why it could be a good idea to go for something along the lines of $1/(x(\ln x)^a)$. The logarithm might not seem like an obvious choice, unless you have already encountered the integrals of this type. But I bet that you have, since such functions are very popular calculus exercise material.
