= Solution
The <completed Dirichlet L-function> is
$$
\Lambda(s,\chi)=\left(\frac q\pi\right)^{(s+a)/2}\Gamma\left(\frac{s+a}{2}\right)L(s,\chi).
$$
For nonprincipal <primitive Dirichlet characters>, termwise <Mellin transformation> initially in $\Re s>1$ gives
$$
\Lambda(s,\chi)=\frac12\int_0^\infty\theta_\chi(x)x^{(s+a)/2}\,\frac{dx}{x}.
$$
The <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 obtain
$$
\boxed{\Lambda(s,\chi)=\varepsilon_\chi\Lambda(1-s,\overline\chi).}
$$
The 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 $0,-2,-4,\ldots$ for a nonprincipal even character, and $-1,-3,-5,\ldots$ 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.
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