Is there a known asymptotic for $A(X):= \sum_{1 \leq i,j \leq X} \frac{1}{\mathrm{lcm}(i,j)}$?

Question

My guess is that there exists a constant $C$ such that $A(X) \sim C (\log X)^2$.

Answer 1

$$ \sum_{1 \leq i,j \leq X} \frac{1}{\mathrm{lcm}(i,j)} = \sum_{1 \leq i,j \leq X} \frac{\mathrm{gcd}(i,j)}{ij} $$ $$ = \sum_{1 \leq i,j \leq X} \frac{\sum_{d|i,j} \phi(d)}{ij}$$ $$ = \sum_{d \leq X} \phi(d) \sum_{1 \leq i,j \leq X: d|i,j} \frac{1}{ij}$$ $$ = \sum_{d \leq X} \frac{\phi(d)}{d^2} \sum_{1 \leq i',j' \leq X/d} \frac{1}{i'j'}$$ $$ = \sum_{d \leq X} \frac{\phi(d)}{d^2} ( \sum_{1 \leq i \leq X/d} \frac{1}{i})^2$$ $$ = \sum_{d \leq X} \frac{\phi(d)}{d^2} ( \log(X/d) + O(1))^2$$ $$ = \sum_{d \leq X} \frac{\phi(d)}{d^2} ( \log^2(X) - 2 \log(d) \log(X) + \log^2(d) + O( \log X ) )$$ $$ = A_0 \log^2 X - 2 A_1 \log X + A_2 + O( \log^2 X )$$

where $$ A_j := \sum_{d \leq X} \frac{\phi(d) \log^j d}{d^2}.$$

One can compute asymptotics for the $A_j$ by Perron's formula, but we proceed instead by elementary means. Since $\phi(d) = \sum_{d=ab} \mu(a) b$ we have $$ A_j = \sum_{ab \leq X} \frac{\mu(a) b \log^j(ab)}{a^2b^2}$$ $$ = \sum_{a \leq X} \frac{\mu(a)}{a^2} (\frac{\log^{j+1}(X)}{j+1} + O( (1 + \log^j(a)) \log^j X ) )$$ $$ = \frac{1}{j+1} \log^{j+1} X \sum_{a \leq X} \frac{\mu(a)}{a^2} + O(\log^j X)$$ $$ = \frac{1}{(j+1)\zeta(2)} \log^{j+1} X + O(\log^j X).$$ Hence $$ \sum_{1 \leq i,j \leq X} \frac{1}{\mathrm{lcm}(i,j)} = \frac{1}{3 \zeta(2)} \log^3 X + O(\log^2 X).$$

Answer 2

If we are talking about "elementary", then just multiply the original sum by the sum of inverse squares and note that if we have three numbers $a=a'd, b=b'd, n^2$ where $(a',b')=1$, $d,n$ are arbitrary, then $A=a'n, B=b'n, d$ are arbitrary $3$ integers with the product $LCM(a,b)n^2$. The original triple sum has the bounds $a'd, b'd\le X$, $n$ formally unrestricted, but restricting it to $[1,N]$ for fixed $N$ changes the triple sum $1+O(1/N)$ times. But then we can squeeze the bounds on $A,B,d$ between $Ad\le X, Bd\le X$ (treating the implicit $n=(A,B)$ as unrestricted) to get the lower bound and $Ad\le NX, Bd\le NX$ (assuming $n\le N$ now) to get the upper bound, so we get an answer for the triple sum of $\frac 1{ABd}$ between roughly speaking $\frac13\log^3 X$ and $\frac13\log^3(XN)$ with additive error $O(\log^2(XN))$ (no number theory in this sum!), which have the same asymptotic up to $1+O(\frac{\log N}{\log X})$. Taking $N$ about $\sqrt{\log X}$, we get the total multiplicative error $1+O(\frac{1}{\sqrt{\log X}})$, which is, of course, suboptimal but who cares. :-)