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For positive integers $n\geq 2$, let $\operatorname{rad}(n)$ the radical of the integer $n$. For example $\operatorname{rad}(24)=6$.

See this Wikipedia to know this definition.

Question. I would like to know if it is possible to discuss the convergence of this integral $$\int_2^\infty\frac{dx}{\left(\operatorname{rad}\left(\lfloor x\rfloor\right)\right)^{\alpha}\log ^2(x)},\tag{1}$$ where $\lfloor x\rfloor$ is the floor function and $\alpha\geq 1$ a fixed real number. Thanks in advance.

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  • $\begingroup$ Please what is $Rad$? $\endgroup$
    – Guy Fsone
    Oct 6 '17 at 12:13
  • $\begingroup$ I've added the link, but here is the definiton: $\operatorname{rad}(1)=1$, and for integers $n>1$ with prime factorization $n=\prod_{p\mid n}p^{e_p}$ then $\operatorname{rad}(n)=\prod_{p\mid n}p$, thus it is the product of different primes dividing $n$. Is a well known multiplicative function, and famous since this arithmetic function is involved in the abc conjecture. Many thanks @GuyFsone $\endgroup$
    – user243301
    Oct 6 '17 at 12:24
  • $\begingroup$ Do you mean $\log \circ \log$ or $(\log)^2$? $\endgroup$
    – Mr. Chip
    Oct 6 '17 at 15:44
  • $\begingroup$ It is the function $(\log(x))\cdot(\log(x))=(\log(x))^2=\log^2(x)$. Many thanks for your attention. I hope that now it is right, and also I hope that you think different version of this integrals, for different number theoretic functions @Mr.Chip $\endgroup$
    – user243301
    Oct 6 '17 at 18:18
  • $\begingroup$ I'm curious, what motivated your investigation of this integral? $\endgroup$
    – FofX
    Oct 10 '17 at 16:07
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As pointed out by other users, we may study the summation version instead. Let $\mathbb{P} = \{p_1,p_2,\cdots\}$ be the set of primes and $\mathbb{N} = \{1,2,\cdots\}$ be the set of positive integers. Then for $\alpha, \beta > 0$ let us consider the sum

$$ S_{\alpha,\beta} := \sum_{n=2}^{\infty} \frac{1}{(\operatorname{rad}n)^{\alpha} (\log n)^{\beta}} = \frac{1}{\Gamma(\beta)} \sum_{n=2}^{\infty} \frac{1}{(\operatorname{rad}n)^{\alpha}} \int_{0}^{\infty} t^{\beta-1} n^{-t} \, dt. $$

Here, the equality follows from the gamma integral $ \int_{0}^{\infty} t^{\beta-1} e^{-st} \, dt = \Gamma(\beta)s^{-\beta}$ for $\beta, s > 0$. Now we partition $n$'s according to the value of radicals. Then resulting classes can be naturally indexed by non-empty finite subsets $I$ of $\mathbb{N}$ in the following way.

$$ \bigg\{ n \in \mathbb{N} : \operatorname{rad}(n) = \prod_{i\in I} p_i \bigg\} = \bigg\{ \prod_{i\in I} p_i^{k_i} : (k_i)_{i\in I} \in \mathbb{N}^I \bigg\} $$

Using this partition, we find that

\begin{align*} S_{\alpha,\beta} &= \frac{1}{\Gamma(\beta)} \sum_{\substack{I \subset \mathbb{N} \\ 0<|I|<\infty}} \sum_{(k_i)_{i\in I} \in \mathbb{N}^{I}} \int_{0}^{\infty} t^{\beta-1} \left( \prod_{i\in I} p_i^{-\alpha-k_i t} \right) \, dt \\ &= \frac{1}{\Gamma(\beta)} \sum_{\substack{I \subset \mathbb{N} \\ 0<|I|<\infty}} \int_{0}^{\infty} t^{\beta-1} \left( \prod_{i\in I} \frac{1}{p_i^{\alpha}(p_i^t - 1)} \right) \, dt \tag{1} \\ &= \frac{1}{\Gamma(\beta)} \int_{0}^{\infty} t^{\beta-1} \left[ \prod_{i=1}^{\infty} \left( 1 + \frac{1}{p_i^{\alpha}(p_i^t - 1)} \right) - 1 \right] \, dt. \tag{2} \end{align*}

Here, interchanging the order of summations and integrals is justified by the Tonelli's theorem. Also we utilized geometric series formula to obtain $\text{(1)}$.

Now using the inequality $e^x - 1 \leq x e^x$, we find that each integral in $\text{(1)}$ satisfies

$$ \int_{0}^{\infty} t^{\beta-1} \left( \prod_{i\in I} \frac{1}{p_i^{\alpha}(p_i^t - 1)} \right) \, dt \geq \int_{0}^{\infty} t^{\beta-|I|-1} \left( \prod_{i\in I} \frac{1}{p_i^{\alpha +t} \log p_i} \right) \, dt, $$

which diverses if $|I| \geq \beta$. So the sum diverges for all $\alpha, \beta > 0$ and hence the same is true for the integral:

$$ \int_{2}^{\infty} \frac{dx}{(\operatorname{rad}\lfloor x\rfloor)^{\alpha}(\log x)^{\beta}} = \infty. $$

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  • $\begingroup$ Are you still using Equation (2)? $\endgroup$
    – Hans
    Oct 14 '17 at 19:21
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    $\begingroup$ Nice use of the Gamma function trick. +1 $\endgroup$
    – Hans
    Oct 14 '17 at 19:58
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    $\begingroup$ Many thanks, I am going to study your answer. $\endgroup$
    – user243301
    Oct 14 '17 at 20:08
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    $\begingroup$ Nice answer and very clever use of $\Gamma$. +1. I wish I could see why our answers disagree, though... Please let me know if you can. $\endgroup$
    – Mr. Chip
    Oct 14 '17 at 21:54
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    $\begingroup$ @Mr.Chip, I failed to see how $d^{\beta} \leq n < d^{\beta+1}$ is related to the bound $n = p_1^{\beta_1}\cdots p_m^{\beta_m} \geq d^{\beta}$ so that your counting argument works. Probably this is where our answers begin to disagree. $\endgroup$ Oct 14 '17 at 22:02
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We have $\int \limits_{2}^{\infty} \frac{dx }{(rad(\lfloor x\rfloor)^\alpha \ln^2 x} \geq \sum \limits_{n=2}^{\infty} \frac{1}{rad(n)^\alpha} \int \limits_{n}^{n+1} \frac{dx}{\ln^2 x} \geq \sum \limits_{n=2}^{\infty} \frac{1}{rad(n)^\alpha \ln^2(n+1)} \geq \sum \limits_{n=2}^{\infty} \frac{1}{rad(n)^\alpha \ln^2(n^2)} = \sum \limits_{n=2}^{\infty} \frac{1}{4 rad(n)^\alpha \ln^2(n)}$

Now i am interested in numbers of the form $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ where $m \geq e_1,e_2,\cdots ,e_k \geq \lceil \alpha \rceil$.

Its easy to see that $rad(n) = p_k \# = e^{\theta(p_k)}$ where $\theta(x) = \sum \limits_{p \leq x} \ln p$ is the Chebyshev's theta function.

The minimum sum is when all powers equal $m$, also for every prime there are $m-\lceil \alpha \rceil$ choices for the power, so there are $(m-\lceil \alpha \rceil)^{\pi(p_k)}$ in total,

So we arrive at $ \sum \limits_{n=2}^{\infty} \frac{1}{4 rad(n)^\alpha \ln^2(n)} \geq \sum \limits_{p_k \geq 2} \frac{(m-\lceil \alpha \rceil)^{\pi(p_k)}}{4 (e^{\theta(p_k)})^{\alpha} \ln^2((p_k \#)^m)}$ for some integer $m$ very large.

$\sum \limits_{p_k \geq 2} \frac{(m-\lceil \alpha \rceil)^{\pi(p_k)}}{4 (e^{\theta(p_k)})^{\alpha} \ln^2((p_k \#)^m)} \geq \sum \limits_{p_k \geq 17} \frac{(m-\lceil \alpha \rceil)^{\frac{p_k}{\ln p_k}}}{4 e^{\alpha \theta(p_k)} m^2 \ln^2((p_k \#))} = \sum \limits_{p_k \geq 17} \frac{(m-\lceil \alpha \rceil)^{\frac{p_k}{\ln p_k}}}{4 e^{\alpha \theta(p_k)} m^2 \ln^2((p_k \#))} =\sum \limits_{p_k \geq 17} \frac{(m-\lceil \alpha \rceil)^{\frac{p_k}{\ln p_k}}}{4 e^{\alpha \theta(p_k)} m^2 \theta^2(p_k)} $,

We know that $\theta(x) < 1.1 x$ for all $x \geq 1$, so

$\sum \limits_{p_k \geq 17} \frac{(m-\lceil \alpha \rceil)^{\frac{p_k}{\ln p_k}}}{4 e^{\alpha \theta(p_k)} m^2 \theta^2(p_k)} \geq \sum \limits_{p_k \geq 17} \frac{(m-\lceil \alpha \rceil)^{\frac{p_k}{\ln p_k}}}{4 e^{1.1 \alpha p_k} *m^2 1.1^2 p_k^2} \geq \sum \limits_{p_k \geq 17} \frac{(m-\lceil \alpha \rceil)^{\frac{p_k}{\ln p_k}}}{5 e^{1.1 \alpha p_k} *m^2 p_k^2} $

Assume that $\lceil \alpha \rceil < 0.5m$ since $\alpha$ is fixed value, so we get that

$\sum \limits_{p_k \geq 17} \frac{(0.5m)^{\frac{p_k}{\ln p_k}}}{5 e^{1.1 \alpha p_k} *m^2 p_k^2} $

Let $m=e^{\ln^2 p_k}$, so we get that $ \sum \limits_{p_k \geq 17} \frac{(0.5m)^{\frac{p_k}{\ln p_k}}}{5 e^{1.1 \alpha p_k} *m^2 p_k^2} = \sum \limits_{p_k \geq 17} \frac{(0.5e^{\ln^2 p_k})^{\frac{p_k}{\ln p_k}}}{5 e^{1.1 \alpha p_k} *(e^{\ln^2 p_k})^2 p_k^2} \geq \sum \limits_{p_k \geq 17} \frac{e^{p_k \ln p_k}}{5 p_k^2 2^{p_k} e^{1.1 \alpha p_k+2 \ln^2 p_k} }$

We have $5 p_k^2 2^{p_k} < e^{p_k}$ for all $p_k \geq 29$ (Proof is easy by induction on all integers).

We also have that $2 \ln^2 p_k < p_k$ for all $ p_k \geq 17$

So we end up with $ \sum \limits_{p_k \geq 29} \frac{e^{p_k \ln p_k}}{e^{1.1 \alpha p_k +p_k+p_k} } = \sum \limits_{p_k \geq 29} \frac{e^{p_k \ln p_k}}{e^{(1.1 \alpha +2) p_k} } = \sum \limits_{p_k \geq 29} e^{p_k \ln p_k-(1.1 \alpha +2) p_k}$

So when $ p_k \ln p_k-(1.1 \alpha +2) p_k \geq 0$ we know that $e^{p_k \ln p_k-(1.1 \alpha +2) p_k} \geq 1$,

And this is true when $\ln p_k \geq 2+1.1 \alpha$, and since $\alpha \geq 1$ so $2\alpha \geq 2$ , so $\ln p_k \geq 3.1 \alpha \geq 2+1.1 \alpha $

Which is true for all $p_k \geq \alpha^{3.1}$ thus we have

$ \sum \limits_{p_k \geq Max(\alpha^{3.1},29)} 1 = +\infty$ since there are infinitely many primes.

thus concluding the proof, the integral diverges

Note : intuitively the number of numbers $n$ that are way bigger than $rad(n) \ln^2 n$ is dense and since the harmonic series and harmonic integral diverges one would suspect that this integral also diverges.

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  • $\begingroup$ Many thanks, I am going to study your answer. $\endgroup$
    – user243301
    Oct 14 '17 at 20:08

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