 Solution

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