= Solution
Let $a_0$ be the constant <Fourier coefficient> and suppose first that $k>0$ is even and $M_k\ne0$; thus $k\geq4$. Put $\varepsilon=i^k=(-1)^{k/2}$. The precise <Mellin continuation of a noncuspidal modular form> is
$$
\boxed{\Lambda_f(s)=(2\pi)^{-s}\Gamma(s)L(f,s),\qquad
\Lambda_f(s)=\varepsilon\Lambda_f(k-s).}
$$
It is <meromorphic> on the whole plane, with possible simple <poles> only at $0,k$, of residues $-a_0,\varepsilon a_0$, respectively. For a <cusp form> it is an <entire function>. A nonzero constant term makes both of these poles genuine.
We first justify the initial <Mellin transform>. Decompose $f=a_0E_k+h$ with $h\in S_k$. For a <cusp form>, $y^{k/2}|h(x+iy)|$ is invariant under the <modular group> and bounded on its fundamental domain, since it tends to zero at the cusp. Therefore it is bounded throughout the <complex upper half-plane>. Integrating the <Fourier coefficient> at height $y=1/n$ gives $|a_n(h)|\leq Ce^{2\pi}n^{k/2}$. The <Eisenstein series> coefficients satisfy $\sigma_{k-1}(n)\leq\zeta(k-1)n^{k-1}$ by replacing divisors $d$ by $n/d$. Consequently $a_n(f)=O(n^{k-1})$, and absolute termwise integration is valid for $\Re s>k$:
$$
\Lambda_f(s)=\int_0^\infty(f(iy)-a_0)y^{s-1}\,dy.
$$
Indeed $\int_0^\infty e^{-2\pi ny}y^{s-1}\,dy=(2\pi n)^{-s}\Gamma(s)$, and the sum of the absolute integrals converges in that half-plane.
Write $I_f(s)=\int_1^\infty(f(iy)-a_0)y^{s-1}\,dy$. Exponential cusp decay of the difference implies uniform convergence on compact subsets of the $s$-plane, including after any number of derivatives in $s$, so $I_f$ is entire. The modular transformation $f(i/y)=(iy)^kf(iy)$ gives
$$
f(iy)=\varepsilon y^{-k}f(i/y).
$$
On the lower half of the Mellin integral, subtract the transformed constant explicitly:
$$
f(iy)-a_0=\varepsilon y^{-k}(f(i/y)-a_0)+a_0(\varepsilon y^{-k}-1).
$$
Substitute $y=1/t$ in the decaying term and integrate the two powers in the constant term. This yields, initially for $\Re s>k$,
$$
\boxed{\Lambda_f(s)=I_f(s)+\varepsilon I_f(k-s)+a_0\left(\frac{\varepsilon}{s-k}-\frac1s\right).}
$$
The right side supplies the claimed <meromorphic continuation> and its residues. Replacing $s$ by $k-s$ and multiplying by $\varepsilon$, with $\varepsilon^2=1$, leaves this expression unchanged, proving the <functional equation> with its correct sign. In particular the residue at $k$ is tied to the residue at zero by that equation; subtracting only $a_0$ at infinity and overlooking its transformed value at zero would miss these poles.
In weight zero every <modular form> is constant, so $L(f,s)=0$ and its completion is identically zero. The same holds for the zero form in odd, negative, or weight-two spaces. Thus the assertion covers all weights in the notation of the paper. There is no conjugation of $f$ in this level-one functional equation.
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