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Expand the two Dirichlet polynomials, keeping the complex conjugate on $B$. The diagonal $m=n$ contributes exactly $T{\sum }_{n\le X}{a}_n\overline{b_n}/n^{2\sigma }$. If $m\ne n$, the exponential integral iswhose absolute value is at most $2/|\log (m/n)|$. Thus the mean-value formula isThis also covers $T=0$.

For $m>n$, integration of $1/x$ over $[n,m]$ gives $\log (m/n)\ge (m-n)/m$; the analogous inequality holds with $m,n$ exchanged. Hence $|\log (m/n){|}^{-1}\le \max (m,n)/|m-n|\le X$. This gives the requested first bound $O(X{\sum }_{m\ne n}|a_n||b_m|/(n^{\sigma }{m}^{\sigma }))$.

For the weighted bound, put ${\alpha }_n=|a_n|n^{-\sigma }$ and ${\beta }_m=|b_m|m^{-\sigma }$. The inequalityreduces the estimate to a row sum and its symmetric counterpart. For a fixed $n$, the preceding logarithm bounds giveHere $n^2/(m(n-m))=n(1/m+1/(n-m))$, and each resulting harmonic sum is $O(\log (2X))$. The symmetric row sum is bounded by $O(m\log (2X))$. Therefore the second error bound is

Expand the finite Euler product into geometric series. By unique prime factorization, it is the sum of $1/n$ over positive integers whose prime factors do not exceed $x$, and hence contains every term up to $m=\lfloor x\rfloor$. The harmonic sum satisfiesThe last inequality uses $m+1>x$, including when $x$ is an integer. For each prime number $p$,The telescoping series over all integers $n\ge 2$ bounds the sum of these remainders by $\frac{1}{2}{\sum }_{n\ge 2}[n(n-1){]}^{-1}=1/2$. Taking logarithms of the product bound therefore givesThis explicit constant suffices; no asymptotic theorem about the distribution of prime numbers is needed.

Writeuniformly for $j\ge 1$. The accumulated error from $j=n+1$ to $2n$ tends to zero, while the corresponding harmonic sum tends to $\log 2$. Hence $b_n\to \log 2$.