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Inspired by this question Is there a known asymptotic for $A(x):=\sum_{1\leq i,j \leq X} \frac{1}{\text{lcm}(i,j)}$? I tried to find the asymptotic of the following function. $$ \Lambda(x)=\sum_{\substack{ 1 \leq m,n \leq x \\ \text{gcd}(m,n)=1}} \frac{1}{mn}. $$ My approach: $$ \left(\sum_{1\leq k \leq x} \frac{1}{k}\right)^2=\sum_{1\leq l \leq x} \frac{\Lambda\big(\frac{x}{l}\big)}{l^2}\label{1}\tag{1} $$ Now, $$ f(x)=\left(\sum_{1\leq k \leq x} \frac{1}{k}\right)^2≈(\ln(x)+\gamma)^2 $$ From, \eqref{1} can establish the approximate identity
$$ 2f(x)-\Lambda(x)≈ 2\int_{1}^{x} \frac{\Lambda(\frac{x}{t})}{t^2} dt \label{2}\tag{2}$$ or, $$ 2f(x)-\Lambda(x)≈ \frac{2}{x}\int_{1}^{x} {\Lambda(\varphi)} d\varphi $$ Using the Newton-Leibniz rule we get

$$x\Lambda'(x)+3\Lambda(x)≈4(\ln(x)+\gamma)+2(\ln(x)+\gamma)^2$$

Solving this differential equation we get, $$ \Lambda(x)≈\frac{2}{3}\ln^2(x)+\left(\frac{8}{9}+\frac{4}{3}\gamma\right)\ln(x)+\left(\frac{2}{3}\gamma^2+\frac{8}{9}\gamma-\frac{8}{27}\right)+\frac{c_1}{x^3} $$ ($c_1$ is the integral constant, for large $x$ this term can be neglected).

My question: Is the asymptotic formula correct? If not, then how to find the asymptotic of the function $\Lambda(x)$?

Is the method correct?

Edit: Though the answer comes wrong with the relation \eqref{2} , but if we use the identity involving the equation $A(x)$ instead of ${\zeta_x}^2(1)=\tau(x)$, then we get the correct answer (the leading term). The approximation \eqref{2} works well here. See my answer below.

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  • $\begingroup$ Experimentally (with Mathematica, defining \[CapitalLambda][x_]:=Sum[1/Times@@mn,{mn,Select[Tuples[Range[x],2],CoprimeQ@@#&]}])$$x^3\left(\Lambda(x)-\left(\frac{2}{3}\ln^2(x)+\left(\frac{8}{9}+\frac{4}{3}\gamma\right)\ln(x)+\left(\frac{2}{3}\gamma^2+\frac{8}{9}\gamma-\frac{8}{27}\right)\right)\right)$$goes like$$0.561097,0.72963,-1.77106,-22.5681,-41.879,-140.577,-202.573,-384.69,-575.653,-962.996,-1180.86,-1858.96,-2198.3,-3077.97,-3986.3,-5207.03,-5898.68,-7809.84,-8719.26,-10982.,-13018.1,-15792.3,-17282.4,-21169.4,-23506.3,...$$ Does not look like this will tend to a finite limit $c_1$ $\endgroup$ Aug 20 '20 at 5:42
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    $\begingroup$ @მამუკაჯიბლაძე: The asymptotic formula in the original post is incorrect, even the leading term is incorrect. For the correct asymptotics, see my post. $\endgroup$
    – GH from MO
    Aug 20 '20 at 7:08
  • $\begingroup$ @GHfromMO I see, thanks. Would it be difficult to determine the constant in your $O$ term? $\endgroup$ Aug 20 '20 at 9:11
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    $\begingroup$ @მამუკაჯიბლაძე: I updated my post to include the second main term. It is $C\log x$, where $C=1.3947995\dots$. $\endgroup$
    – GH from MO
    Aug 20 '20 at 10:50
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    $\begingroup$ @GHfromMO Spectacular! Thanks $\endgroup$ Aug 20 '20 at 12:46
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We have, for $x\geq 2$, \begin{align*} \sum_{\substack{ 1 \leq m,n \leq x \\ \mathrm{gcd}(m,n)=1}} \frac{1}{mn} &=\sum_{1 \leq m,n \leq x}\frac{1}{mn}\sum_{k\mid\mathrm{gcd}(m,n)}\mu(k)\\ &=\sum_{1\leq k\leq x}\mu(k)\sum_{\substack{ 1 \leq m,n \leq x \\ k\mid\mathrm{gcd}(m,n)}} \frac{1}{mn}\\ &=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\sum_{1\leq m\leq x/k}\frac{1}{m}\right)^2\\ &=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log\frac{x}{k}+\gamma+O\left(\frac{k}{x}\right)\right)^2\\ &=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log^2\frac{x}{k}+2\gamma\log\frac{x}{k}+O(1)\right)\\ &=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}\left(\log^2 x-2\log x\log k+2\gamma\log x+O(\log^2 k)\right)\\ &=S_1(x)(\log^2 x+2\gamma\log x)-S_2(x)(2\log x)+O(1), \end{align*} where \begin{align*} S_1(x)&:=\sum_{1\leq k\leq x}\frac{\mu(k)}{k^2}=\frac{6}{\pi^2}+O\left(\frac{1}{x}\right),\\ S_2(x)&:=\sum_{1\leq k\leq x}\frac{\mu(k)\log k}{k^2}=\frac{36\zeta'(2)}{\pi^4}+O\left(\frac{\log x}{x}\right). \end{align*}

We conclude that, for $x\geq 2$, $$\sum_{\substack{ 1 \leq m,n \leq x \\ \mathrm{gcd}(m,n)=1}} \frac{1}{mn}= \frac{6}{\pi^2}\log^2 x+C\log x+O(1),$$ where $$C:=\frac{12\gamma}{\pi^2}-\frac{72\zeta'(2)}{\pi^4}=1.3947995\dots$$

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    $\begingroup$ One can also replace the first $O(1)$ by $O(k\log x/x).$ It will give the constant term in asymptotic formula and the error term $O(\log^2 x/x).$ $\endgroup$ Feb 1 at 8:59
  • $\begingroup$ @AlexeyUstinov: Indeed, since $\sum_{k=1}^\infty\mu(k)\log^2 k/k^2$ converges. $\endgroup$
    – GH from MO
    Feb 1 at 9:25

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