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The completed Dirichlet L-function isFor nonprincipal primitive Dirichlet characters, termwise Mellin transformation initially in $\Re s>1$ givesThe Dirichlet character theta function decays exponentially at infinity; its transformation makes it decay faster than any power at zero. Hence the integral is entire in $s$. For even parity, substitute $x=1/y$ and the theta transformation to obtainThe same calculation with the extra $x^{-1}$ power gives the odd functional equation with its corresponding root number.

The gamma function has no zeros and has simple poles at nonpositive integers. Thus the nontrivial zeros of $L$ and $\Lambda$ coincide with multiplicities. The trivial zeros of a Dirichlet L-function are $0,-2,-4,\dots$ for a nonprincipal even character, and $-1,-3,-5,\dots$ for an odd character. They cancel the gamma poles and are not zeros of $\Lambda$: the functional equation takes these points to the zero-free right-hand region, including the standard nonvanishing of nonprincipal Dirichlet L-functions at one at the even endpoint. The canceled zeros are simple.

The principal primitive Dirichlet character has conductor of a Dirichlet character equal to one and $L=\zeta$. In that case $\Lambda ={\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ is meromorphic with poles at zero and one. Its canceled trivial zeros begin at $-2$, while $\zeta (0)=-1/2$ is not zero. Multiplication by $s(s-1)/2$ produces the entire Riemann xi function used below.

The Riemann xi function is the entire functionwith $\xi (s)=\xi (1-s)$. Its Hadamard factorization iswhere 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 $|s|\le 2T$, functional symmetry reduces to $\Re s\ge 1/2$. 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. ThusFor a zero with $|\rho |\le T$, its contribution in Jensen's formula on radius $2T$ is at least $\log 2$. ConsequentlyIf 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 $-\frac{1}{2}\log z$. The correct expansion is $(z-\frac{1}{2})\log z-z+\frac{1}{2}\log (2\pi )+O(|z{|}^{-1})$ in a fixed sector. Its consequence $\log |\Gamma (z)|=O(|z|\log (2+|z|))$ is all that the argument needs.