What is the polynomial approximation for Riemann zeta function in the strip since it is holomorphic?

Question

Probably it is known that Holomorphic function behave like polynomial with large degree and Riemann zeta function is meromorphic continuation such that we know that Every holomorphic function is meromorphic, but not vice versa, And Holmorphic function can be approximated by Taylor expansion theorme on the disk or by Weierstrass factorisation theorem then it is good to get polynomial approximation of Riemann zeta function in the strip at a least to solve many problems related for example by compositional inverse of that zeta which it is very interesting in complex analysis ,Any way for approximating Riemann zeta function in the strip ?

Answer 1

Let $$F_n(s) = \int_{-\infty}^\infty \zeta(i x) n e^{-\pi n^2 (-is-x)^2}dx$$

It is entire and as $n \to \infty$ it converges locally uniformly to $\zeta(s)$ on $\Re(s) < 1$.

Proof : That is it entire is obvious. With the change of variable $x = t-y$ for $t$ real we have $$F_n(it) = \int_{-\infty}^\infty \zeta(i (t-y)) n e^{-\pi n^2 y^2}dy$$ thus by analytic continuation for $ \Re(s) < 1$ $$F_n(s) = \int_{-\infty}^\infty \zeta(s-iy) n e^{-\pi n^2 y^2}dy$$ which (due to the continuity and the polynomial growth of $\zeta(s)$ on vertical strips) converges locally uniformly to $\zeta(s)$ on $\Re(s) < 1$.

Let $$F_{n,K_n}(s) = \sum_{k=0}^{K_n} \frac{F_n^{(k)}(0)}{k!} s^k$$ where $K_n$ is chosen such that $|F_{n,K_n}(s)-F_n(s)| \le 1/n$ on $\Re(s) \in [-n,1-1/n] , \Im(s) \in [-n,n]$, then $F_{n,K_n}(s) \to \zeta(s)$ locally uniformly on $\Re(s) <1$.