Let $f=\sum a(n)q^n\in S_k(N,\chi)$ be a cusp form of integral weight. Can someone give me the proof of the fact that : the Hecke $L$-function $L(f,s)=\sum\frac{a(n)}{n^s}$ is entire.
I searched in Koblitz's book "Introduction to Elliptic Curves and Modular forms" and Diamond's book "A First Course in Modular forms" but I cannot find anything about it.
The short answer is very straightforward. I'll let you fill in the details.
As $f$ is cuspidal, you have that $f(iy) \to 0$ super-polynomially fast as $y \to \infty$. Consider the integral function $$ I(s) = \int_0^\infty f(iy) y^s \frac{dy}{y}.$$ The super-decay of $f(iy)$ guarantees convergence for every $s$, and this is clearly an analytic function in $s$. For $s$ with $\text{Re} s$ large enough, we may substitute the Fourier expansion to recognize $$ I(s) = \int_0^\infty \sum a(n) e^{-2\pi ny} y^s \frac{dy}{y}.$$ As $$ \int_0^\infty e^{-2\pi n y} y^s \frac{dy}{y} = (2\pi n)^{-s} \Gamma(s),$$ we see that $$ I(s) = L(s,f) \Gamma(s) (2\pi)^{-s}.$$ So $L(s)$ is necessarily entire, and $I(s)$ is really the completed $L$-function $\Lambda(s)$.
Futher, one can prove the functional equation by using the modularity condition satisfied by $f(iy)$ under the action of the distinguished matrix $\left( \begin{smallmatrix} & -1 \\ N& \end{smallmatrix}\right)$.