I am reading Serre's article ( Formes modulaires et fonctions zêta p-adiques ).
At some point, it is written that for a modular form $f$, we can find $a_0(f)$ in terms of $a_n(f)'s$.
The procedure to do that is as follows :
Let $\phi_f(s) = \sum_{n=1}^{\infty} a_n n^{-s}$ and then this $\phi_f(s)$ extends to a meromorphic function and the value at $s=0$ gives the constant term.
Since $\phi_f(s)$ is literally the $L$-function, then it has a functional equation. Using this functional equation
For cusp form, because of pole of Gamma function, it becomes clear to me that corresponding $L$-function will vanish.
For Eisenstein series, $L$-function factors as product of two other $L$-functions and then the constant term is also clear.
However, how to prove that the statement above is true in general? For a modular form which is a "mixture" of Eisenstein and cusp, why the above procedure works?
UPDATE : More precisely, I want to ask why $a_0(f)$ is equal to $L(f,s)|_{s=0}$?
From the Laplace transform identity
$$ \int_0^\infty x^{a-1}e^{-bx}\mathrm dx={\Gamma(a)\over b^a}, $$
we have
\begin{aligned} \phi_f(s) &={(2\pi)^s\over\Gamma(s)}\int_0^\infty x^{a-1}\left(\sum_{n\ge1}a_n e^{-2\pi nx}\right)\mathrm dx \\ &={(2\pi)^s\over\Gamma(s)}\int_0^\infty x^{a-1}(f(ix)-a_0)\mathrm dx. \end{aligned}
When $f$ is a modular form of $SL_2(\mathbb Z)$ of even weight $k$, we have
$$ f(ix)=(-1)^{k/2}x^{-k}f(i/x). $$
By splitting the integral into $\int_0^1+\int_1^\infty$ and perform appropriate substitutions (details in section 6.16 of Apostol's Modular Functions and Dirichlet Series in Number Theory), we have
\begin{aligned} (2\pi)^{-s}\Gamma(s)\phi_f(s) &=\int_1^\infty(f(ix)-a_0)(x^s+(-1)^{k/2}x^{k-s}){\mathrm dx\over x} \\ &-a_0\left(\frac1s+{(-1)^{k/2}\over k-s}\right). \end{aligned}
Observe that the integral on the right hand side is entire, so we have as $s\to0$ that
$$ \phi_f(s)\sim-{a_0\over s\Gamma(s)}=-{a_0\over\Gamma(s+1)}\to-a_0 $$