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Use the Fourier transform convention $\hat{m}(t)=\int e^{-itx}dm(x)$. The complex measure is finite because ${\sum }_n|f(n)|n^{-\sigma }<\infty$. The Möbius divisor-sum identity, proved in the preceding solution, allows multiplication of absolutely convergent Dirichlet series and givesHere $\zeta$ is the Riemann zeta function. We prove the estimates in the intended range $1<\sigma \le 2$. A uniform estimate of the printed form for all $\sigma >1$ would be false: at $t=0$, $F(\sigma )\to 1$ as $\sigma \to \infty$, while its proposed upper bound tends to zero. A version valid for every $\sigma >1$ replaces $(\sigma -1{)}^{-3/4}$ by $1+(\sigma -1{)}^{-3/4}$.

For completeness, unique prime factorization and absolute convergence give $\zeta (s)={\prod }_p(1-p^{-s}{)}^{-1}$ andas the logarithm defined by that Euler product. The elementary nonnegative trigonometric polynomialtherefore proves the three-four-one inequality for Euler productsWe also need elementary uniform bounds for derivatives of the Riemann zeta function. For an integer $H\ge 1$, summation by parts givesThis continues the function to $\Re s>0$ with just the pole at one. For $1\le \sigma \le 2$, $|t|\ge 1$, choose $H=\lceil 2+|t|\rceil$. Differentiating $j$ times, the finite Dirichlet polynomial is bounded by ${\sum }_{n\le H}(\log n{)}^j/n{\ll }_j{\log }^{j+1}(2+|t|)$. Derivatives of the second term have the same bound, since $|s-1|\ge 1$. In the last term, differentiated integrands are bounded by $u^{-\sigma -1}(\log u{)}^j$, whose integral is $O_j(H^{-\sigma }{\log }^j(2H))$; multiplication by $|s|\ll H$ is harmless. It follows thatAs $\zeta (\sigma )\le 1+(\sigma -1{)}^{-1}$, the three-four-one inequality for Euler products gives, for $|t|\ge 1$,To handle bounded $t$, first note that $\zeta$ cannot vanish at $1+i{t}_0$ for $t_0\ne 0$: a zero of order $m\ge 1$ would make the three-four-one product $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound one. At $s=1$, the reciprocal $F$ extends holomorphically with a simple zero, since the pole of $\zeta$ has residue one. Thus $F$ and all its derivatives are bounded on the compact rectangle $1\le \sigma \le 2$, $|t|\le 1$. This proves the requested reciprocal zeta bounds near the line one:

For the derivative estimate, repeatedly differentiate $F=1/\zeta$. Every term of $F^{(k)}$ is a constant timesThe preceding derivative bounds and reciprocal estimate bound this by $O_k((\sigma -1{)}^{-3(k+1)/4}{\log }^{3k+1}(2+|t|))$. Compact-height bounds supply the remaining range. Since ${\partial }_t^{k}F(\sigma +it)=i^k{F}^{(k)}(\sigma +it)$, this has exactly the requested dependence, with an explicit logarithmic exponent linear in $k$ for $k\ge 1$.

Here is the unsmoothing a logarithmically weighted sum argument with the normalization repaired. For real $v$,Indeed the Fourier transform of $u{e}^{-\sigma u}{1}_{u\ge 0}$ is $(\sigma +it{)}^{-2}$, so Fourier inversion gives the identity, also at $v=0$ by continuity. In a vertical contour integral the coefficient is $1/(2\pi i)$ and the differential is $ds=idt$. The printed real-$t$ integral has $dt$ with $1/(2\pi i)$ and is missing this factor of $i$.

Set $a_m=\mu (m)(\log m{)}^k$, taking $(\log 1{)}^0=1$ when $k=0$. Its Dirichlet series is $(-1{)}^k{F}^{(k)}(s)$. The logarithmically smoothed Perron formula, justified by absolute convergence, givesFor $x\ge 3$ choose $\sigma =1+1/\log x$. The integral of ${\log }^{3k+1}(2+|t|)/|\sigma +it{|}^2$ is finite, with a constant depending only on $k$, and $x^{\sigma }=ex$. Thus, with $\beta =3(k+1)/4$,For $0<h\le 1$, subtraction gives the exact finite-difference identityThe last sum is $O_k((x{h}^2+h)(\log x{)}^k)$, since there are $O(xh+1)$ terms. ConsequentlyFor $k\ge 3$, choose $h=(\log x{)}^{(\beta -k)/2}$ to get exponent $(\beta +k)/2=7k/8+3/8$. For $0\le k<3$, the trivial bound $O_k(x(\log x{)}^k)$ is already no larger than that estimate. The extra $(\log x{)}^k$ term is absorbed for fixed $k$. HenceThis also yields the cancellation needed for the Prime number theorem. Choose a fixed $k>3$, discard $m\le \sqrt{x}$ at cost $O(\sqrt{x})$ for the unweighted sum, and use partial summation with $(\log m{)}^{-k}$ on the rest. The displayed bound gives $M(x)=O_k(x(\log x{)}^{-k/8+3/8})=o(x)$; the preceding solution then gives $\psi (x)\sim x$.