# entire functions with prescribed zeros

we have two entire functions $f^{+}(s)$ & $f^{-}(s)$ that have the following symmetries :

$$f^{+}(s)-f^{+}(-s)=\pi s\cos(\pi s)$$
$$f^{-}(s)-f^{-}(-s)=-\pi s\cos(\pi s)$$ $$f^{+}(s)=f^{-}(-s)$$
and have the prescribed zeros:

$$f^{+}(s)=0$$,

$s=0,1,2,3...$

$$f^{-}(s)=0$$,

$s=0,-1,-2,-3...$

can we use Weierstrass factorization theorem to express the two functions in terms of infinite products !?

-

Yes, you can (if the nonnegative integers are the only zeros of $f^{+}$). It will tell you that $f^{+}(s) = e^{h(s)} s \prod_{n=1}^\infty (1 - s/n) e^{s/n}$ for some entire function $h$. If there could be other zeros as well, all you can say is $f^{+}(s) = g(s) s \prod_{n=1}^\infty (1 - s/n) e^{s/n}$ for some entire function $g$.
By the way, $f^{+}(s) - f^{+}(-s) = \pi s \cos(\pi s)$ says that $f^{+}(s) = \frac{\pi}{2} s \cos(\pi s) + e(s)$ where $e(s)$ is an arbitrary even entire function. Then $f^{-}(s) = f^{+}(-s) = - \frac{\pi}{2} s \cos(\pi s) + e(s)$. The prescribed zeros say $e(s) = \frac{\pi s}{2} (-1)^s$ for nonnegative integers $s$.
EDIT: That product $$\prod_{n=1}^\infty (1-s/n) e^{s/n} = \dfrac{e^{\gamma s}}{\Gamma(1-s)}$$ where the right side is interpreted as $0$ at the positive integers (which are poles of $\Gamma(1-s)$).