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A Dirichlet polynomial is a finite Dirichlet series, $A(s)={\sum }_{n\le X}{a}_n{n}^{-s}$. Vertical integration separates equal product indices from oscillatory unequal ones.

The Riemann zeta function satisfies ${\int }_0^T|\zeta (1/2+it){|}^{4}dt\ll T(\log T{)}^4$ for $T\ge 2$. Squaring a short Dirichlet polynomial produces the divisor function; the divisor-square summatory bound controls the off-diagonal and diagonal terms.

For finite Dirichlet polynomials, ${\int }_T^{2T}A(\sigma +it)\overline{B(\sigma +it)}dt$ has diagonal term $T{\sum }_n{a}_n\overline{b_n}/n^{2\sigma }$. Its off-diagonal error is bounded by ${\sum }_{m\ne n}|a_n{b}_m|/(n^{\sigma }{m}^{\sigma }|\log (m/n)|)$. A weighted row-sum estimate further bounds this by $\log (2X){\sum }_{n\le X}{n}^{1-2\sigma }(|a_n{|}^2+|b_n{|}^2)$.

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 $\psi (x)\sim x$.

We use the Van der Corput sum-integral lemma. Put $e(u)=e^{2\pi iu}$ and $I_h={\int }_a^{b}e(f(x)-hx)dx$. The Fourier series of the periodization of $1_{[a,b]}(x)e(f(x))$ giveswhere integer endpoints have half weight. This is the Dirichlet-Jordan convergence theorem for a piecewise smooth, or more generally bounded-variation, periodic function. Here $f$ is $C^1$, so the periodized function has bounded variation. Changing to the requested endpoint convention costs at most one.

Write $u=f^{\prime }$. For $h\ne 0$, $|u-h|\ge |h|-\delta >0$. Since $u$ is continuous and monotone, the reciprocal has bounded variation, and integration by parts in the Riemann-Stieltjes sense yieldsThe variation of the reciprocal is at most $2\delta /(h^2-{\delta }^2)$. Summing over $h\ne 0$ gives $O((1-\delta {)}^{-1})$, separating $|h|=1$ and using convergence of ${\sum }_{h\ge 2}{h}^{-2}$. For each endpoint, useThe symmetric partial sums of the first term are a constant multiple of ${\sum }_{h=1}^H\sin (2\pi hx)/h$, uniformly bounded in $H$ and $x$; this standard Fourier series bound follows by splitting at $h\asymp 1/\Vert x\Vert$ and applying Abel summation to the remaining sine sum. The second term is absolutely summable with bound $O((1-\delta {)}^{-1})$. The same bound therefore holds for the whole sum of the $h\ne 0$ integrals. Since $I_0$ is the ordinary integral,No second derivative is required; monotonicity supplies the needed variation estimate.

For the Hardy-Littlewood approximation to the Riemann zeta function, take $f(w)=-t\log w/(2\pi )$. On $w\ge x\ge |t|/\pi$, $|f^{\prime }(w)|\le 1/2$ and $f^{\prime }$ is monotone. The proved lemma says that the difference between the partial sum of $w^{-it}$ and its integral over $(x,Y]$ is $O(1)$ uniformly in $Y$. Weighted Abel summation with the decreasing weight $w^{-\sigma }$ then makes the weighted difference $O(x^{-\sigma })$, since its total variation on $[x,\infty )$ is $x^{-\sigma }$. Initially for $\sigma >1$, the tail integral is $x^{1-s}/(s-1)$. The bounded primitive of the discrepancy gives a locally uniformly convergent weighted discrepancy integral for every $\sigma >0$, continuing the identity to that region. Thus, away from the pole,If $t=0$ the ordinary sum-integral comparison supplies the same estimate. At $s=1$ the formula is understood meromorphically. It approximates the Riemann zeta function by a finite Dirichlet polynomial, transfers exponential sum estimates to bounds in the critical strip, yields elementary near-one bounds for $\zeta$ and its derivative, and supports estimates for the mean value of Dirichlet polynomials and numerical calculations.

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

Let $D(s)={\sum }_{p\le X}{p}^{-s}$, a Dirichlet polynomial supported on primes. The cosine sum is $(D(\sigma +it)+\overline{D(\sigma +it)})/2$, so its $j$th power is a finite linear combination of $D(\sigma +it{)}^k\overline{D(\sigma +it{)}^{j-k}}$, $0\le k\le j$.

The Dirichlet polynomial $D(s{)}^k$ is supported on integers that are products of exactly $k$ primes, counted with multiplicity. Each coefficient is at most $k!$ by unique prime factorization; for $k=0$ the only coefficient is the one at $1$. As $j$ is odd, $k\ne j-k$, so no integer can appear in both supports. Thus every diagonal term vanishes in the mean value of Dirichlet polynomials from part (a).

Both supports lie in $[1,X^j]$. The first error bound in part (a), with that common length, now gives for each termSumming the finitely many terms proves the odd moment of a prime cosine sum estimate:The argument holds for all real $\sigma$ and all $T\ge 0$; no cancellation estimate involving $T$ is needed once the diagonal is absent.

Put $P(t)={\sum }_{n\le \sqrt{t/(2\pi )}}{n}^{-1/2-it}$. The given bound for the Riemann zeta function, together with $(a+b{)}^4\le 8(a^4+b^4)$ for $a,b\ge 0$, givesThe integral over $[0,1]$ is bounded, since the Riemann zeta function has no pole on this compact segment. It remains to estimate the fourth moment of the Riemann zeta function through this moving Dirichlet polynomial cutoff.

Let $L=\lfloor \sqrt{T/(2\pi )}\rfloor$ and $Y=L^2$. In the expansion of $|P(t){|}^4$, a quadruple $n_1,n_2,n_3,n_4\le L$ occurs precisely forIts oscillatory factor is $e^{it\log (n_3{n}_4/(n_1{n}_2))}$, and its weight is $(n_1{n}_2{n}_3{n}_4{)}^{-1/2}$. When $n_1{n}_2=n_3{n}_4$, its integral has length at most $T$. When these products differ, its integral over $[t_0,T]$ has absolute value at most $2/|\log (n_3{n}_4/(n_1{n}_2))|$. Thus the moving cutoff affects the lower endpoint but preserves the off-diagonal bound from part (a).

Write $r_L(m)=\#\{(a,b):a,b\le L,ab=m\}$. The divisor function $d(m)$ bounds $r_L(m)$. Grouping the off-diagonal majorant by the two products and using the weighted row-sum argument in part (a), with coefficient $r_L(m)$ and $\sigma =1/2$, bounds it byThis uses only the given divisor-square summatory bound. For bounded $T$, the desired conclusions follow by boundedness on compact segments, so these estimates may be read for large $T$ with $Y\ge 2$. We have proved the requested diagonal-plus-error estimate:

The diagonal sum equals ${\sum }_{m\le Y}{r}_L(m{)}^2/m\le {\sum }_{m\le Y}d(m{)}^2/m$. If $D(x)={\sum }_{m\le x}d(m{)}^2\ll x{\log }^3(2x)$, then partial summation givesConsequently the full fourth-moment bound is