A smoothed von Mangoldt explicit formula

Write $\mathrm{\Phi }=\tilde{\varphi }$. Integration by parts gives $\mathrm{\Phi }(s)=\frac{y^{s+1}-x^{s+1}}{(y-x)s(s+1)}\mathrm{.}$ Its only pole is $0$, with residue $1$; the apparent singularity at $-1$ is removable, with value $-\log (y\mathrm{/}x)\mathrm{/}(y-x)$. On each fixed vertical strip it is $O_{x,y}(\mathrm{\mid }\mathrm{\Im }s{\mathrm{\mid }}^{-2})$ at large height, with the constant also depending on the strip.

We compute the constant at zero. Put $\gamma ={\lim }_N(H_N-\log N)$ as in The Euler–Mascheroni constant and the harmonic asymptotic. The fractional-part formula in For $\mathrm{Re}s>0$, zeta admits the fractional-part integral formula with a simple residue-one pole at $1$ and ${\int }_1^N\frac{\{u\}}{u^2}du=\log N-H_N+1$ give $\zeta (1-s)=-1\mathrm{/}s+\gamma +O(s)$. Logarithmic differentiation of the locally uniform product in The Weierstrass product for reciprocal Gamma at $1$ gives $-\frac{{\mathrm{\Gamma }}^{\mathrm{\prime }}(1)}{\mathrm{\Gamma }(1)}=1+\gamma +{\sum }_{n\ge 1}(\frac{1}{n+1}-\frac{1}{n})=\gamma \mathrm{.}$ The same product at $1$ gives $\mathrm{\Gamma }(1)=1$, so $\mathrm{\Gamma }(1-s)=1+\gamma s+O(s^2)$. Finally The Riemann zeta function satisfies the classical sine-gamma functional equation and $2^s{\pi }^{s-1}\sin (\pi s\mathrm{/}2)=\frac{s}{2}(1+s\log (2\pi )+O(s^2))$ yield $\zeta (s)=-1\mathrm{/}2-s\log (2\pi )\mathrm{/}2+O(s^2)$: the two Euler constants cancel. Thus ${\zeta }^{\mathrm{\prime }}(0)\mathrm{/}\zeta (0)=\log (2\pi )$.

To justify inversion explicitly, let $J(z)=\frac{1}{2\pi i}{\int }_{\mathrm{\Re }s=2}\frac{z^s}{s(s+1)}ds(z>0)\mathrm{.}$ Closing a rectangle to the left for $z>1$, and to the right for $z<1$, gives $J(z)=1-z^{-1}$ and $J(z)=0$, respectively, by The residue theorem for a null-homologous cycle. Indeed, first let the height tend to infinity with the other vertical side fixed: the horizontal integrals are $O(T^{-2})$ times a fixed width. Then let that side tend to the appropriate infinity through half-integers; its integral is $O(z^{\sigma }\mathrm{/}\mathrm{\mid }\sigma \mathrm{\mid })$ and vanishes. At $z=1$, continuity of the absolutely convergent initial integral gives $J(1)=0$. Consequently $[yJ(y\mathrm{/}u)-xJ(x\mathrm{/}u)]\mathrm{/}(y-x)=\varphi (u)$ for every $u>0$, including $u=x,y$. Combining this with step 1.1 and The logarithmic derivative of the zeta Dirichlet series is the Dirichlet series of the von Mangoldt function on Re s greater than 1 gives ${\sum }_n\mathrm{\Lambda }(n)\varphi (n)=\frac{1}{2\pi i}{\int }_{\mathrm{\Re }s=2}-\frac{{\zeta }^{\mathrm{\prime }}}{\zeta }(s)\mathrm{\Phi }(s)ds\mathrm{.}$ The exchange of sum and integral is absolute, since ${\sum }_n\mathrm{\Lambda }(n)n^{-2}\le {\sum }_{n\ge 2}(\log n)n^{-2}<\mathrm{\infty }$ and $\mathrm{\Phi }(2+it)=O_{x,y}((1+\mathrm{\mid }t\mathrm{\mid }{)}^{-2})$.

By [L1], [L2], and $0<\mathrm{\Re }\rho <1$, step 1.1 gives absolute convergence of the zero sum: its bands at large $\mathrm{\mid }\mathrm{\Im }\rho \mathrm{\mid }\asymp j$ contribute $O_{x,y}(\log (j+2)\mathrm{/}{j}^2)$. The trivial-zero sum converges absolutely since $\mathrm{\mid }\mathrm{\Phi }(-2k)\mathrm{\mid }\le C_{x,y}{x}^{1-2k}\mathrm{/}{k}^2$. We also choose admissible heights explicitly. Let $M_j$ count zeros with ordinates in $[j-1,j+2]$, with multiplicity. It is $O(\log (j+2))$. Among the $2M_j+2$ equally spaced points of $[j,j+1]$, each such ordinate excludes at most one point at distance less than $1\mathrm{/}(4M_j+4)$. Choose the least remaining point $T_j$. Ordinates outside that larger interval are at distance at least $1$, so every zero ordinate is at distance at least $1\mathrm{/}(4M_j+4)$ from $T_j$. Conjugation gives the same separation at $-T_j$.

Fix an odd integer $R\ge 3$ and shift the integral of step 2.1 to $\mathrm{\Re }s=-R$, using heights $\pm T_j$ from step 2.2. On $-1\le \mathrm{\Re }s\le 2$, A local formula for the logarithmic derivative of zeta bounds ${\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta$ by $O({\log }^2{T}_j)$: there are $O(\log T_j)$ nearby zeros, each reciprocal is $O(\log T_j)$, and the pole term is bounded. On $-R\le \mathrm{\Re }s\le -1$, A left-half-plane bound for the logarithmic derivative of zeta gives $O_R(\log T_j)$. Thus each horizontal integral is $O_{R,x,y}({\log }^2{T}_j\mathrm{/}{T}_j^2)$ and tends to zero. By Residues in the von Mangoldt contour shift, $-{\zeta }^{\mathrm{\prime }}\mathrm{/}\zeta$ has residues $1$ at $1$, $-m$ at a zero of multiplicity $m$, and $-1$ at each trivial zero. Multiplication by $\mathrm{\Phi }$ therefore gives residues $\mathrm{\Phi }(1)$, $-m\mathrm{\Phi }(\rho )$, $-\mathrm{\Phi }(-2k)$, and, by step 1.2, $-\log (2\pi )$ at $0$. There is no pole at $-1$. The residue theorem and absolute convergence in step 2.2 now express the initial integral as these residues for $-2k>-R$, the full nontrivial-zero sum, and the upward integral on $\mathrm{\Re }s=-R$.

On that last line, the distance to every trivial zero is at least $1$. Step 1.1 gives $\mathrm{\mid }\mathrm{\Phi }(-R+it)\mathrm{\mid }\le C_{x,y}\frac{x^{1-R}}{R^2+t^2}\mathrm{.}$ The left-half-plane bound therefore makes its integral at most $C_{x,y}{x}^{1-R}{\int }_{\mathbb{R}}\frac{\log (R+\mathrm{\mid }t\mathrm{\mid }+2)}{R^2+t^2}dt\le C_{x,y}^{\mathrm{\prime }}{x}^{1-R}\frac{\log (R+2)}{R}⟶0.$ Letting odd $R$ tend to infinity in step 3.1, using $x>1$ and the absolute convergence from step 2.2, proves the formula for every stated pair $1<x<y$.