= Solution
With the normalization of the preceding completion, the super-completion is the <Riemann xi function>
$$
\boxed{\xi(s)=\frac12s(s-1)\Xi(s)
=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s).}
$$
The <pole-subtracted theta integral for the completed zeta function> gives the particularly useful expression
$$
\xi(s)=\frac12+\frac14s(s-1)\bigl(J(s)+J(1-s)\bigr).
$$
It proves directly that $\xi$ is an <entire function>, with $\xi(0)=\xi(1)=1/2$ and $\xi(s)=\xi(1-s)$. Here “integral function” means entire, not integer-valued.
To prove the <order-one growth of the Riemann xi function>, put $M(R)=\max_{|s|\leq R}|\xi(s)|$. For $R\geq3$, the exponentially decreasing theta integrand bounds both $J(s)$ and $J(1-s)$ by
$$
C\int_1^\infty e^{-\pi u}u^{(R+1)/2-1}\,du
\leq C\pi^{-(R+1)/2}\Gamma((R+1)/2).
$$
Multiplication by the quadratic factor in $s$ and the <Stirling formula> imply $\log M(R)=O(R\log R)$. Hence the <order of an entire function> is at most one. On the positive real axis, $\zeta(R)=1+O(2^{-R})$, and the defining formula and <Stirling formula> give
$$
\log\xi(R)=\frac R2\log R+O(R).
$$
This lower bound proves that the order is at least one. Consequently \b[$\xi$ is entire of order exactly one], generally not of finite exponential type.
Its zeros are precisely the <Nontrivial zeros of the Riemann zeta function>, counted with multiplicity. The values at zero and one are nonzero; the <Gamma function> has no zeros; and the completed <functional equation> together with the <Euler product> shows that there are no further completed zeros outside the closed critical strip. By <Hadamard factorization> for entire functions of order one,
$$
\xi(s)=e^{A+Bs}\prod_\rho\left(1-\frac{s}{\rho}\right)e^{s/\rho},
\qquad e^A=\frac12,\quad B=\frac{\xi'(0)}{\xi(0)}.
$$
The genus-one factors give $\sum_\rho|\rho|^{-2}<\infty$, and the <logarithmic derivative> converges locally normally away from the zeros:
$$
\frac{\xi'(s)}{\xi(s)}
=B+\sum_\rho\left(\frac1{s-\rho}+\frac1\rho\right).
$$
Differentiate the defining product of $s(s-1)$, the gamma factor, and $\zeta(s)$. Rearranging proves the <global partial-fraction expansion of the zeta logarithmic derivative>:
$$
\boxed{\frac{\zeta'(s)}{\zeta(s)}
=B-\frac1s-\frac1{s-1}+\frac12\log\pi
-\frac12\frac{\Gamma'(s/2)}{\Gamma(s/2)}
+\sum_\rho\left(\frac1{s-\rho}+\frac1\rho\right).}
$$
The summands must be kept together: each is $O_s(|\rho|^{-2})$ for large $|\rho|$, whereas separating the two complex reciprocal sums need not preserve <absolute convergence>. The gamma term accounts for the trivial zeros as well as cancellation at zero.
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