= Solution
The <Riemann xi function> is the <entire function>
$$
\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s),\qquad \xi(0)=\xi(1)=\frac12,
$$
with $\xi(s)=\xi(1-s)$. Its <Hadamard factorization> is
$$
\xi(s)=\frac12e^{Bs}\prod_{\rho}\left(1-\frac{s}{\rho}\right)e^{s/\rho},
\qquad B=\frac{\xi'(0)}{\xi(0)},
$$
where the <Nontrivial zeros of the Riemann zeta function> are repeated by multiplicity and the factors are canonical genus-one factors.
Here is the growth estimate needed for the <Jensen zero-count bound>. For $|s|\le2T$, functional symmetry reduces to $\Re s\ge1/2$. Euler summation truncated at $T^2$ bounds $(s-1)\zeta(s)$ by a fixed power of $T$, uniformly in that region; the multiplication cancels the <pole> at one. The logarithmic gamma estimate bounds $\log|\Gamma(s/2)|$ by $O(T\log T)$, including the bounded small-$s$ part separately. The remaining elementary factors obey the same bound. Thus
$$
\max_{|s|\le2T}\log|\xi(s)|\le C T\log T.
$$
For a zero with $|\rho|\le T$, its contribution in <Jensen's formula> on radius $2T$ is at least $\log2$. Consequently
$$
n_\xi(T)\log2\le\frac1{2\pi}\int_0^{2\pi}\log|\xi(2Te^{i\theta})|\,d\theta-\log|\xi(0)|
\ll T\log T.
$$
If a zero lies on the integration circle, use nearby radii and <continuity> of the zero-count estimate. Hence $n_\xi(T)=O(T\log T)$ for $T>2$. The growth also gives order at most one and justifies the stated <Hadamard factorization>; the zero-count bound gives convergence of its genus-one factors.
The printed logarithmic Stirling hint drops the term $-\tfrac12\log z$. The correct expansion is $(z-\tfrac12)\log z-z+\tfrac12\log(2\pi)+O(|z|^{-1})$ in a fixed sector. Its consequence $\log|\Gamma(z)|=O(|z|\log(2+|z|))$ is all that the argument needs.
Back to article page
