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 [ 11 , Theorem 1.3 ] .

Follows directly from the definitions of Σ₁ and Σ₂.

See [ 27 , 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 [ 23 ]

This follows from Theorem 10.6.1. TODO: create a primary Buthe section to place this result

Multiplying both sides of ?? 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 ?? with \(b_0 = \log B\). Combining these bounds, we derive ??.

By [ 4 , (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 [ 4 , (1.5) ] to obtain the claim.

This is [ 3 , 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 [ 6 , 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 \(w(0) = ∌u\), 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.

TBD.

TBD.

TBD.

TBD.