This question has been bugging me for the past few days. I suspect that it is too difficult for my level of maths (3rd year undergrad). I have enjoyed struggling and wrestling with it but now I must get back to exam revision...I give up. I think I might be close, but every time I attempt to write an answer I confuse myself... anyway here's the problem:

Does there exist a function, $g(x)$ say, such that for any given $c$ in $\mathbb{R}$, $g(x)$ is eventually (strictly) LESS THAN $c/x$, but the area under $g(x)$ is infinite? The "whole point" is that, if no such $g(x)$ exists, then $1/x$ really is on the "borderline" up to a constant, and this question gets rid of annoying functions such as $a(1/x-1/x^2)$ which have infinite area... but let's be honest, $a(1/x-1/x^2)$ tends to $1/x$ as $x$ tends to infinity, so really this question gets rid of such functions which obstruct (what would be) our aim.

I suspect the answer to the question is no. Here is my latest effort:

For each $n\epsilon\mathbb{N}$, there must be at least one $a\in\mathbb{R}$ such that eventually

$$ ax^{-1-1/n}<f(x)<\frac{a}{x}\tag{*} $$

Fix $n\epsilon\mathbb{N}$ . Then if $\{a_i\}$ is a (possibly uncountable) set of real numbers satisfying $(*)$ .

Let $A_n=\inf\{a_i\}$ .

Now, $(*)$ must hold for every $n$.

This means $a(n)x^{-1-1/n}<f(x)<\frac{a(n)}{x}$ . Now we look at $a(n)$ . Or maybe $A_n$ . . .

. . . Or do we? I don't know how to make progress.

Now all this looks like it might be on the right lines. But I've no idea tbh and at this point in my solution attempt, my brain hurts . . . any help please?

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    $\begingroup$ $1/(x\ln x)$... $\endgroup$ – David Mitra Apr 15 '12 at 19:19

Take for example $g(x)=\dfrac{1}{x\log x}$. An antiderivative is $\log\log x$, which "blows up," albeit at a glacial pace, as $x\to\infty$.

And for something that is in the long run less than $\dfrac{c}{x\log x}$ for any positive constant $c$, take $\dfrac{1}{x(\log x)(\log\log x)}$. It has $\log\log\log x$ as an antiderivative.

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    $\begingroup$ Thanks lol. I have a lot to learn $\endgroup$ – Adam Rubinson Apr 15 '12 at 19:28
  • $\begingroup$ Actually that function is quite amazing. I was inputting values for 1/((x)log(x)) - 1/(x)^1.01, thinking for a moment that after the initial root, there were no more, meaning that 1/((x)log(x)) < 1/(x)^1.01 after this root. But amazingly enough there is a second root at x ~ 1.285 x 10^281. So this function f(x) = 1/(xlogx) is "slow", but eventually cuts through all the 1/x^i, i>1. It cuts through 1/x^i reasonably close to x=1 for large i, and cuts through 1/x^i very far away for i close to but greater than 1. $\endgroup$ – Adam Rubinson Apr 15 '12 at 19:53
  • $\begingroup$ @AdamRubinson: It is an interesting function. I was using it only for its asymptotic (large enough $x$) behaviour. $\endgroup$ – André Nicolas Apr 15 '12 at 19:58

$\int \frac{1}{x\ln (x)}=\ln (\ln (x))+C$, which diverges as $x$ tends to $\infty$.


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