We compute

giving the claim.

The equality is immediate from Theorem 9.2.1 and exchanging the order of \(\sum \) and \(\int \), as is justified by \(\sum _n |\mu (n)|\int _0^{x/n^2} du \leq \sum _n x/n^2 {\lt} \infty \))

This is just splitting the sum at \(K\).

This is just splitting the integral at \(K\), since \(f(u) = M(\sqrt{x/u}) = 0\) for \(x{\gt}u\).

We split into two cases. If \(K{\gt}x\), the second line of 2 is empty, and the first one equals 1, by \(M(t)=0\) for \(t{\lt}1\), so 2 holds.

Now suppose that \(K \leq x\). Then we combine Sublemma 9.2.2 and Sublemma 9.2.3 with Lemma 9.2.1 to give the claim.

TODO.

Routine computation.

See [ 17 , Theorem 1.3 ] .

Follows directly from the definitions of Σ₁ and Σ₂.

See [ 44 , Lemma 2.10 ] .

An argument of Pintz [ s employed. The interval \([0,T]\) is split into subintervals \([T/\lambda ^{k+1}, T/\lambda ^k]\) where \(\lambda {\gt} 1\), \(0 \leq k \leq K-1\), and \(K = \lfloor \frac{\log T/H}{\log \lambda } \rfloor + 1\). Then use the zero-free region to bound \(\Re \rho \).

Inserting (A.6) into the result of (A.12).

Follows from combining Sublemmas ??, ??, ??, and ??.

This is proven by Platt and Trudgian [ 35 ]

This follows from Theorem 10.6.1.

Multiplying both sides of 6 by \(\frac{1}{\sqrt{x}}\) gives

as \(\frac{1}{\sqrt{x}} \leq \frac{1}{e^{\frac{b}{2}}}\). Then, for \(x \geq B\) we apply 7 with \(b_0 = \log B\). Combining these bounds, we derive 8.

By [ 6 , (1.5) ] , (A.27) holds with \(B_0 = 11\), \(B = 10^{19}\), and \(c = 0.94\). Thus we may apply Lemma 9.5.1 with \(B_0 = 11\), \(B = 10^{19}\), and \(c = 0.94\) from [ 6 , (1.5) ] to obtain the claim.

This is [ 5 , Theorem 1 ] .

Values for \(20 \leq b \leq 2000\) are computed using Theorem ??, and values for \(2500 \leq b \leq 25000\) are computed as using Theorem 9.5.1. For \(b {\gt} 25000\) we use Theorem 9.5.2.

From Table 8 we have \(\varepsilon (19 \log 10) = 1.93378 \cdot 10^{-8}\). Now apply Corollary 9.5.1 and Theorem ??.

Use Lemma 3.1.1 and Lemma 3.1.2, similar to the proof of ‘limiting_fourier_aux‘.

Apply Sublemma 9.6.1 and take the limit as \(\sigma \to 1^+\), using the continuity of \(G\) and the dominated convergence theorem, as well as the Fourier inversion formula.

Routine manipulation.

By the nonnegativity of \(a_n\) we have

By Proposition 9.6.1 we can express the right-hand side as

If \(\lambda {\gt} 0\), then \(I_\lambda (y)=0\) for negative \(y\), so

If \(\lambda {\lt} 0\), then \(I_\lambda (y)=e^{-\lambda y}\) for \(y\) negative, so

hence

Since \(x^{-\sigma } * (2\pi x / T) * x^{\sigma -1}/(-\lambda ) = 1/(1-\sigma )\), the result follows.

Similar to the proof of Proposition 9.6.2; see [ 9 , Proposition 2.4 ] for details.

This follows from the definition of \(\Phi ^{\pm ,\circ }_\nu \) and the properties of the \(\coth \) function.

This follows from the definition of \(\Phi ^{\pm ,\circ }_\nu \) and the properties of the \(\coth \) function.

This follows from the definition of \(\Phi ^{\pm ,\circ }_\nu \) and the properties of the \(\coth \) function.

This follows from the definition of \(\Phi ^{\pm ,\circ }_\nu \) and the properties of the \(\coth \) function.

L’Hôpital’s rule can be applied to show the continuity at \(0\).

This follows from the definition of \(B^\pm \) and the fact that \(B^\pm (0) = 1\).

This follows from the definition of \(\Phi ^{\pm ,\ast }_\nu \) and the properties of the \(B^\pm \) function.

This follows from the definition of \(\Phi ^{\pm ,\ast }_\nu \) and the properties of the \(B^\pm \) function.

This follows from the definition of \(\Phi ^{\pm ,\ast }_\nu \) and the properties of the \(B^\pm \) function.

This follows from the definition of \(\Phi ^{\pm ,\ast }_\nu \) and the properties of the \(B^\pm \) function.

The residues cancel out.

Since \(\Phi ^{\pm , \circ }_\nu (z)\) and \(\Phi ^{\pm , \circ }_\nu (z)\) have no poles on \(\mathbb {R}\), they have no poles on some open neighborhood of \([-1,1]\). Hence they are \(C^2\) on this interval. Since \(w(0) = \nu \), we see that \(\Phi ^{\pm , \ast }_\nu (0)=0\), giving the claim.

Since \(\Phi ^{\pm , \circ }_\nu (z)\) and \(\Phi ^{\pm , \circ }_\nu (z)\) have no poles on \(\mathbb {R}\), they have no poles on some open neighborhood of \([-1,1]\). Hence they are \(C^2\) on this interval. Since \(w(0) = \nu \), we see that \(\Phi ^{\pm , \ast }_\nu (0)=0\), giving the claim.

By the preceding lemmas it suffices to verify continuity at \(0, -1, 1\). Continuity at \(0\) is clear. For \(t = -1, 1\), by \(\coth \frac{w(t)}{2} = \coth \frac{\nu }{2}\), we see that \(B^{\pm }(w(t)) = \left(\frac{\nu }{2} - \pi i t\right)\left(\coth \frac{\nu }{2} \pm 1\right)\), and so

hence, by Definition 9.6.6, \(\varphi ^{\pm }_{\nu }(t) = 0\). Thus, \(\varphi ^{\pm }_{\nu }\) is continuous at \(-1\) and at \(1\).

The function \(\coth w = 1 + \frac{2}{e^{2w}-1}\) is bounded away from the imaginary line \(\Re w = 0\), that is, it is bounded on \(\Re w \geq \kappa \) and \(\Re w \leq -\kappa \) for any \(\kappa {\gt} 0\). The map \(w(z) = \nu - 2\pi i z\) sends the line \(\Im z = -\frac{\nu }{2\pi }\) to the imaginary line, and the region \(\Im z \geq c\) is sent to \(\Re w \geq 2\pi c + \nu \).

Similar to previous lemma.

The bound on \(\Phi ^{\pm ,\star }_{\nu }\) follows from the bound on \(\Phi ^{\pm ,\circ }_{\nu }\) by \(\Phi ^{\pm ,\star }(z) = \frac{1}{2\pi i}\bigl(w\, \Phi ^{\pm ,\circ }(w) - \nu \, \Phi ^{\pm ,\circ }(\nu )\bigr)\).

Similar to previous lemma.

For all \(t \neq 0\), by the identities \(2\cosh \frac{t}{2}\sinh \frac{t}{2} = \sinh t\) and \(2\sinh ^2\frac{t}{2} = \cosh t - 1\),

Since \(e^u\) is convex, \(e^u \geq 1 + u\) for all \(u \in \mathbb {R}\). We apply this inequality with \(u = t\) and \(u = -t\) and obtain the conclusion for \(t \neq 0\). Since \(B^{\pm }(t)\) is continuous at \(t = 0\), we are done. .

Similar to previous. .

By the definition of the Fourier transform, and the fact that \(\varphi ^{\pm }_{\nu }\) is supported on \([-1,1]\).

Since \(\Phi ^{\pm ,\circ }_{\nu }(z) \pm \Phi ^{\pm ,\star }_{\nu }(z)\) has no poles in the upper half plane, we can shift contours upwards, as we may: for \(\Im z \to \infty \), \(e(-zx) = e^{-2\pi i z x}\) decays exponentially on \(\Im z\), while, by Lemma 1.3, \(\Phi ^{\pm ,\circ }_{\nu }(z) \pm \Phi ^{\pm ,\star }_{\nu }(z)\) grows at most linearly, and so the contribution of a moving horizontal segment goes to \(0\) as \(\Im z \to \infty \).

This follows from the \(\pi i\)-periodicity of coth.

Follows from previous lemma.

We have \(\Phi ^{\pm ,\circ }_{\nu }(z) - \Phi ^{\pm ,\star }_{\nu }(z) = -\Phi ^{\pm ,\star }_{\nu }(z+1)\) and \(\Phi ^{\pm ,\circ }_{\nu }(z) + \Phi ^{\pm ,\star }_{\nu }(z) = \Phi ^{\pm ,\star }_{\nu }(z-1)\), and so the formula in the previous lemma simplifies to

We would like to integrate along \(\Re z = 0\), but \(\Phi ^{\pm ,\circ }_{\nu }(z)\) has a pole at \(z = -\frac{i\nu }{2\pi }\); when dealing with this issue, we have to take care not to introduce poles on the lines \(\Re z = -1\) and \(\Re z = 1\) by separating \(\Phi ^{\pm ,\circ }_{\nu }\) and \(\Phi ^{\pm ,\star }_{\nu }\) prematurely. As \(\Im z \to -\infty \), \(e(-zx) = e^{-2\pi i z x}\) decays exponentially on \(\Im z\), while, by Lemma 1.3, \(\Phi ^{\pm ,\circ }_{\nu }(z) \pm \Phi ^{\pm ,\star }_{\nu }(z)\) grows at most linearly.

since the pole is at \(-\frac{i\nu }{2\pi }\), the residue of \(\Phi ^{\pm ,\circ }_{\nu }(z)\) at the pole is \(\frac{i}{2\pi }\), and our path goes clockwise. .

Again by Cauchy’s theorem and decay as \(\Im z \to -\infty \)

Similar to previous.

This follows from the previous lemma.

This follows from the previous lemma.

This follows from the symmetries of \(\varphi ^{\pm }_{\nu }\).

Apply Lemmas 9.6.9, 9.6.10, 9.6.11 We know \(\varphi _\nu ^\pm \) is integrable because it is \(C^1\) on \([-1, 0]\) and \([0, 1]\), and identically \(0\) outside \([-1, 1]\)./

By Lemmas 9.6.16, 9.6.17, the integrands in Lemmas 9.6.18, 9.6.19 are non-negative. Hence, the bound holds for all \(x \neq 0\). By definition, \(I_\nu \) is right-continuous. Since \(\varphi _\nu ^\pm \in L^1(\mathbb {R})\), \(\widehat{\varphi _\nu ^\pm }\) is continuous on \(\mathbb {R}\). Thus, letting \(x \to 0^+\), we see that the bound holds for \(x = 0\) as well.

Apply Lemmas 9.6.9, 9.6.10, 9.6.11 We know \((\varphi _\nu ^\pm )'\) is integrable because it is \(C^1\) on \([-1, 0]\) and \([0, 1]\), and identically \(0\) outside \([-1, 1]\)./

Apply Lemmas 9.6.9, 9.6.10, 9.6.11 We know \(\varphi _\nu ^\pm \) is absolutely continuous because it is \(C^1\) on \([-1, 0]\) and \([0, 1]\), and identically \(0\) outside \([-1, 1]\)./

Since \((\varphi _\nu ^\pm )'\) is \(C^1\) on \([-1, 0]\) and on \([0, 1]\), the \(L^1\) norm of \((\varphi _\nu ^\pm )''\) on each of these intervals is finite, and so \((\varphi _\nu ^\pm )'\) has finite total variation on each of them. As \((\varphi _\nu ^\pm )'\) has right and left limits at \(-1\), \(0\) and \(1\), the jumps at those points are finite, and so their contribution to \(\| (\varphi _\nu ^\pm )'\| _{\mathrm{TV}}\) is finite. /

For \(f\) absolutely continuous with \(f, f' \in L^1(\mathbb {R})\), integration by parts gives us that \(\hat{f}(x) = \widehat{f'}(x)/(2\pi i x)\). If \(f' \in L^1(\mathbb {R})\) with \(\| f'\| _{\mathrm{TV}} {\lt} \infty \), then, again by integration by parts, \(|\widehat{f'}(x)| \leq |f'|_{\mathrm{TV}}/(2\pi x)\). We are done by the preceding lemmas.

We know that \(\varphi _\nu ^\pm \) is continuous and in \(L^1(\mathbb {R})\); by Corollary 9.6.2, \(\widehat{\varphi _\nu ^\pm }\) is in \(L^1(\mathbb {R})\). Hence, Fourier inversion holds everywhere, and in particular for \(t = 0\):

By definition, \(\varphi _\nu ^\pm (0) = \Phi _\nu ^{\pm ,\circ }(0)\), and, by definition, \(\Phi _\nu ^{-,\circ }(0) = \frac{1}{e^\nu - 1}\) and \(\Phi _\nu ^{+,\circ }(0) = \frac{1}{1 - e^{-\nu }}\). Thus,

since \(\int _{-\infty }^{\infty } I_\nu (x)\, dx = 1/\nu \). We are done by Corollary 9.6.1.

See previous.

Since \(z\coth (z)\) is regular at \(0\) and an even function, we see that \(f(z) := (z \coth z)'\) and \(f(z)/z\) are regular at \(0\), and hence analytic on the strip \(|\Im z| \leq \frac{\pi }{2}\). We see from \(f(z) = \coth z - z\operatorname {csch}^2 z\) that \(f(z)\) has at most exponential growth as \(\Re z \to \pm \infty \) within the strip. Hence, by Phragmén–Lindelöf, it is enough to verify the inequalities \(|f(z)/z| \leq 1\) for \(\Im z = \pm \frac{\pi }{2}\) and \(|f(z)| \leq 1\) for \(\Im z = \pm \frac{\pi }{4}\); by complex conjugation, it suffices to check them for \(\Im z = \frac{\pi }{2}\) and \(\Im z = \frac{\pi }{4}\).

By the above, \(f(z) = \frac{(\sinh 2z)/2 - z}{\sinh ^2 z}\). Now, for \(z = x + i\frac{\pi }{4}\) with \(x \in \mathbb {R}\), we have \(\sinh 2z = i\cosh 2x\) and \(\sinh ^2 z = -\frac{1}{2} + \frac{i}{2}\sinh 2x\), and so \(|f(z)|^2 = \frac{(\cosh 2x - \pi /2)^2 + 4x^2}{1 + \sinh ^2 2x}\). By \(1 + \sinh ^2 2x = \cosh ^2 2x\),

Since \(\cosh 2x = 1 + 2\sinh ^2 x \geq 1 + 2x^2\), \(\pi {\gt} \frac{\pi ^2}{4}\) and \(2\pi {\gt} 4\), the numerator here is positive. We conclude that \(|f(z)|^2 {\lt} 1\) for \(z = x + i\frac{\pi }{4}\), as was desired.

For \(z = x + i\frac{\pi }{2}\) with \(x \in \mathbb {R}\), we have \(\coth z = \tanh x\) and \(\operatorname {csch}^2 z = -\operatorname {sech}^2 x\). Then \(|f(z)|^2 = (\tanh x + x\operatorname {sech}^2 x)^2 + \left(\frac{\pi }{2}\operatorname {sech}^2 x\right)^2\). Since \(\operatorname {sech}^2 x - 1 = -\tanh ^2 x\), this is equal to

Since \(|z|^2 \geq \frac{\pi ^2}{4} {\gt} 2\), it suffices to show that \(2x\operatorname {csch} x - 2\operatorname {sech} x - \cosh x \leq 0\) for all \(x \in \mathbb {R}\); by parity, it is enough to check all \(x \geq 0\). The statement is then equivalent to \(g(x) = 2x - 2\tanh x - \sinh x \cosh x \leq 0\), since \(\sinh x \geq 0\). That follows from \(g'(x) = 2\tanh ^2 x - \cosh ^2 x - \sinh ^2 x = -2\sinh ^2 x \tanh ^2 x - 1 \leq 0\) (by \(1 - \cosh ^2 x = -\sinh ^2 x\)) and \(g(0) = 0\).

See previous.

The residue theorem on the region of \(R^+\) between \([1, 1+iT]\) and \(C_n^+\).

The residue theorem on the region of \(\overline{R^+}\) between \([1-iT, 1]\) and \(C_n^-\).

The residue theorem on the region of \(R_C\) between \(C_{n,1}^+ \cup C_{n,1}^-\) and the \(\sigma _n\) column.

For the conjugation-antisymmetric integrand \(G^\star x^s\), \(\int _{C_{n,1}^-} = \overline{\int _{C_{n,1}^+}}\), and \(z - \bar z = 2i\, \Im z\).

As \(\sigma _n \to -\infty \) the truncated horizontal segments exhaust the rays of \(C_\infty \); uses boundedness of \(G x_0^s\) on \(\partial R\) and \(x {\gt} x_0\).

The integrand is \(O((x/x_0)^{\sigma _n})\) via boundedness of \(G^\circ x_0^s\) on \(L\), and \((x/x_0)^{\sigma _n} \to 0\) since \(x {\gt} x_0 \geq 1\) and \(\sigma _n \to -\infty \).

Since \(\sigma _n \to -\infty \) and there are finitely many poles in \(S\), for all large \(n\) the set \(\{ \Re s {\gt} \sigma _n\} \) contains every pole of \(f\) in \(S\); meromorphicity on \(S\) makes the residue vanish at non-poles, so the truncated sum is then constant and equals the full residue sum over \(S\).

\(C_{n,1}^+\) differs from \(C\) (truncated at height \(\delta \)) only in its horizontal segment \(1 + i\delta \to \sigma _n + i\delta \), which exhausts the ray \(1 + i\delta \to -\infty + i\delta \), and its vertical segment \(\sigma _n + i\delta \to \sigma _n + iT\), which vanishes — both as in 9.6.19, 9.6.20, here at height \(\delta \), using boundedness of \(G^\star x_0^s\) on \(L\) and on \(C\).

Assemble from the sub-lemmas. Split the central line into its upper half \([1,1+iT]\) and lower half \([1-iT,1]\), and apply Lemmas 9.6.15 and 9.6.16 to rewrite each as the truncated contour \(C_n^+\) (resp. \(C_n^-\)) plus the residues of \(G\) over \(R^+ \cap \{ \Re s {\gt} \sigma _n\} \) (resp. \(\overline{R^+} \cap \{ \Re s {\gt} \sigma _n\} \)). Split each \(C_n^{\pm }\) into its horizontal \(\Im s = \pm T\) segment and the remainder \(C_{n,1}^{\pm }\). On \(C_{n,1}^{\pm }\) substitute \(G = G^\circ + \mathrm{sgn}(\Im s)\, G^\star \): by 9.6.17 the \(G^\circ \) part becomes the \(\sigma _n\) column plus the residues of \(G^\circ \) over \(R_C \cap \{ \Re s {\gt} \sigma _n\} \), and by 9.6.18 the \(G^\star \) part combines into \(2i\, \Im \int _{C_{n,1}^+} G^\star x^s\). Now let \(n \to \infty \): the \(\Im s = \pm T\) segments converge to \(C_\infty \) (9.6.19); the \(\sigma _n\) column vanishes (9.6.20); \(C_{n,1}^+ \to C\) (9.6.22), so \(\Im \int _{C_{n,1}^+} G^\star x^s \to \Im \int _C G^\star x^s\); the off-axis truncated sums converge to the full (finite) residue sums over \(R^+ \sqcup \overline{R^+} = R \setminus R_C\) (9.6.21), while the \(R_C\) truncated sum converges to the improper residue sum by definition. Collecting terms, and using \(\frac{1}{2\pi i} \cdot 2i = \frac{1}{\pi }\), yields the claim.

Apply Lemma 9.6.25. The \(G^\star \) reflection is the conjugation symmetry of \(\Phi ^\star \) together with \(F(\bar s) = \overline{F(s)}\); boundedness follows from \(\Phi ^\circ \) bounded and \(\Phi ^\star = O(|z|)\) (CH2 Lemma 4.3).

\(|\Phi ^\varepsilon _\lambda (\pm 1 + ir')| \leq |r'|\) (CH2 Lemma 4.3), \(|x^s| = x^{\Re s}\).

‘intC‘ is linear, \(|\mathrm{sgn}(\lambda )| = 1\), and \(|\Im w| \leq |w|\).

By 9.6.23 the left side equals the \(C_\infty \) integral, the \(\frac{1}{\pi } \Im \int _C G^\star \) term, and the two residue sums; subtracting the residue sums (which match exactly) and applying the triangle inequality with 9.6.24 and 9.6.25 gives the \(\frac{1}{2\pi } O^*(E)\) bound.

TBD.

TBD.

TBD.

TBD.