$$\xi(s) = s(s-1)\pi^{-s/2}\Gamma(s/2) \zeta(s)$$ $$\xi(s) = \xi(1-s)$$

thus $\Xi(s) = \xi(1/2+s) = \Xi(-s)$ is even, and furthermore it is an "entire and even Laplace transform" :

$$\Xi(s) = \int_{-\infty}^\infty e^{su} g(u) du$$

with : $$g(u) = \sum_{n=1}^\infty 2 \pi (3 n^2e^{5u/2}-2\pi n^4e^{9u/2})e^{-\pi n^2 e^{2u}}$$

my question is then :

  • Firstly, why these $\Xi(s)$ and $g(u)$ explicit formulas are so ugly ? do they have simpler forms (I'm not sure the Theta series version of wiki/Riemann_zeta_function is simpler) ?

  • What are the theorems that we know about entire and even Laplace transforms ?

  • I think $g(u)$ should be even itself, which is not obvious at all there. $g(u)$ should be also more than exponentially decreasing such that $\Xi(s)$ is holomorphic on the whole complex plane : $g(u) = \mathcal{O}(e^{-k|u|})$ for all $k$ (we can compare $g(u)$ with the derivatives of $\frac{1}{ e^{\scriptstyle e^u}-1}$).

  • $\Xi(s)$ is also special under the Riemann hypothesis because all its zeros are of real part $=0$. do we know many such entire and even Laplace transforms with all its zeros on the vertical line $\Re(s)=0$ ? one of those would be $2 \cosh(s) = \int e^{su} (\delta(u-1)+\delta(u+1))du$. but as you can see, $\delta(u-1)+\delta(u+1)$ is different from $g(u)$ because $g(u)$ is holomorphic :

  • $\Xi(s)$ is then special again because it is the Laplace transform of an holomorphic function. $\sqrt{\pi} e^{\textstyle s^2} = \int_{-\infty}^\infty e^{su} e^{-u^2}du$ is I guess the prototypal of such class of functions : "entire and even Laplace transforms of holomorphic functions". (but $\int_0^\infty e^{su} e^{-u^2}du$ wouldn't belong to that class because $e^{-u^2} 1_{u \ge 0}$ is discontinuous at $u = 0$).

$\Xi(s)$ is entire because it is a product of meromorphic functions, and the poles of $\Gamma(s/2)$ are cancelled by the zeros of $s \zeta(s)$, and $\zeta(s) (s-1)$ is entire. it is even because of the functionnal equation for zeta and the duplication and the complement formula for gamma. and it is a Laplace transform because it is a product of Laplace transforms :

$\pi^{-s/2} \Gamma(s/2) = \int_0^\infty y^{s-1} e^{-\pi y^2} dy$ $ s (s-1) \pi^{-s/2}\Gamma(s/2) = -2\pi(s-1)\int_0^\infty y^{s+1} e^{-\pi y^2} dy = 2 \pi \int_0^\infty y^{s-1} \frac{d}{dy} y^3 e^{-\pi y^2} dy$

and with $f(y) = \frac{d}{dy} y^3 e^{-\pi y^2} = (3y^2-2\pi y^4)e^{-\pi y^2}$ : $$\xi(s) = s (s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s) = 2\pi \int_0^\infty y^{s-1} \sum_{n=1}^\infty f(ny) dy$$ which is of course a Laplace transform. $\Xi(s) = \xi(1/2+s) = 2\pi \int_{-\infty}^\infty e^{su} e^{u/2}\sum_{n=1}^\infty f(ne^u) du = \int_{-\infty}^\infty e^{su} g(u) du$ with $g(u) = 2\pi \sum_{n=1}^\infty e^{u/2} f(ne^u) = \sum_{n=1}^\infty 2 \pi (3 n^2e^{2u}-2\pi n^4e^{4u}) e^{u/2} e^{-\pi n^2 e^{2u}}$



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