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We know that, for positive monotonically increasing functions like $f(x) = \sqrt{x}$ and $f(x) = x$, if you integrate their reciprocals to infinity, it doesn't converge. On the other end, for functions like $f(x)=x^2$, in integral of the reciprocal does converge.

I kind of just assumed that the reciprocal integral will converge iff $f(x)\in\omega(x)$. For those unfamiliar with little-omega notation, basically it means that $f(x)$ asymptotically grows strictly faster than $x$ (i.e. $\lim_{x\to\infty} \frac{x}{f(x)} = 0$).

However, this is wrong. $f(x) = x \cdot \ln(x)$ will also diverge, despite being in $\omega(x)$. And $f(x) = x \cdot \ln^2(x)$ will converge. But $f(x) = x \cdot \ln(x) \cdot \ln(\ln(x))$ won't converge.

So the line between convergence is some weird function that I'm not sure of. Does anyone know what the exact function is that separates convergence and divergence of the reciprocal integral?

If I had to guess, the integral will converge for all functions $f(x)$ that are in $\omega(g(x))$, where $$g(x) = x \cdot \ln(x) \cdot \ln(\ln(x)) \cdot \ln(\ln(\ln(x))) \cdot \cdots$$

I have no idea if that's right, but if it is, I have no idea what the name for that $g(x)$ is, or if there's a better candidate for a function that behaves just like that.

Edit: Added that fact that the functions should be monotonically increasing.

chausies
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  • You may want to add the assumption that the functions $1/f$ involved are monotone decreasing, otherwise examples like a long string of widely spaced narrow spikes also has the property that $\int_1^\infty f$ converges, though $\limsup_{x\to\infty} 1/f(x) = \infty$. – Alex Ortiz Apr 12 '24 at 18:35
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    There is no slowest growing $f$ for which the integral converges, or fastest growing $f$ for which it diverges. An analogous theorem applies to series. You can have necessary conditions, or sufficient ones, but not necessary and sufficient ones based on comparison to any specific function. – J.G. Apr 12 '24 at 18:44
  • @J.G. I think that post answers the question. Thanks! – chausies Apr 12 '24 at 19:00
  • @J.G. I think that comment deserves to be promoted to answer. – leonbloy Apr 16 '24 at 16:32
  • @leonbloy You could argue if pointing to another question doing the same thing for series answers this integral question then it shouldn't be answered, but marked as a duplicate. – J.G. Apr 16 '24 at 16:44

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