# zero distribution of the Fourier kernel $\Phi(u)$ for Riemann $\Xi(z)$ function

Riemann $\Xi(z)$ function is related to Riemann $\zeta(s)$ function via ($s=1/2+i z$):

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

The functional equation for $\zeta(s)$ is equivalent to $\Xi(z)=\Xi(-z)$.

Riemann $\Xi(z)$ function can be expressed as a Fourier transformation:

$$\Xi(z)=2\int_0^{\infty}\Phi(u)\cos(z u){\rm d}u$$

where $$\Phi(u)=\sum_{n=1}^{\infty}\left(4\pi^2n^4\exp(9u/2)-6\pi n^2\exp(5u/2)\right)\exp\left(-\pi n^2 \exp(2u)\right)=\Phi(-u)$$

For more details please refer to the excellent review paper by Dimitrov and Rusev[1]. ([1]: Dimitrov and Rusev, The zeros of entire Fourier transforms, EAST JOURNAL ON APPROXIMATIONS Volume 17, Number 1 (2011), 1-108)

When $\text{Im}(u)\to \frac{\pi}{4}+0^{-}$,$\Phi(u)\to 0$. Numerical results showed that $\Phi(u)$ also have zeros on imaginary axis (i.e. $\text{Re}(u)=0$).

Question 1: When $u\in \mathbb{C}$, is $\Phi(u)$ an entire function?. If yes, what is its order and type?

Question 2: Are there any other zeros that are not in the categories above?

Any references and comments are welcome.

Best regards- mike

$$\kappa(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s) = \int_0^\infty x^{s-1} \sum_{n=1}^\infty e^{- \pi n^2 x^2} dx$$

(it is what Mike calls the the Gamma-completed Riemann zeta function. it is useful because $\kappa(s) = \kappa(1-s)$)

you know that $\displaystyle f(x) = \sum_{n=1}^\infty e^{- \pi n^2 x} = \sum_{m=1}^\infty \frac{(-1)^{d_p(m)}}{e^{\pi mx}-1}$ where $(-1)^{d_p(n)}$ is the coefficient appearing in the Dirichlet series for $\zeta(2s)/\zeta(s)$.

now notice that $\displaystyle \frac{1}{e^x-1} - \frac{1}{e^{2x}-1} = \frac{1}{e^{x}-e^{-x}}$, thus $\displaystyle f(x) - f(2x) = \sum_{m=1}^\infty \frac{(-1)^{d_p(m)}}{e^{\pi mx}-e^{-\pi mx}}$.

define $F(u) = f(e^{2u}) = \sum_{n=1}^\infty e^{- \pi n^2 e^{2u}}$. you get that :

$$\kappa(s) (1-2^{-s/2}) = \int_{-\infty}^\infty e^{su} (f(e^{2u}) - f(2e^{2u}) )du = \int_{-\infty}^\infty e^{su} ( F(u) - F(u + \sqrt{2})) du$$

and because $e^{2(u+i \pi)} = e^{2u}$, $F(u)$ is $i \pi$ perdiocic, so that :

$$\kappa(s) (1-2^{-s/2}) (1-e^{ i \pi s}) = \oint_C e^{su} ( F(u) - F(u + \sqrt{2})) du$$ where $C$ is a the union of two horizontal lines in the complex plane : one going from $-\infty$ to $+\infty$, and the other going from $+\infty+i\pi$ to $-\infty+i\pi$. note that for now, $C$ is not a closed contour yet. we have to prove that $\oint_{+\infty \to +\infty+i \pi} + \oint_{-\infty+i\pi \to -\infty} e^{su} ( F(u) - F(u + \sqrt{2})) du = 0$ in order to close the contour.

when $\Re(e^u) \ne 0$ : $\displaystyle \frac{1}{e^{\pi m e^u}-e^{-\pi m e^u}} \to 0$ when $\Re(u) \to \pm \infty$, so that $\displaystyle F(u) = \sum_{m=1}^\infty \frac{(-1)^{d_p(m)}}{e^{\pi m e^u}-e^{-\pi m e^u}} \to 0$ too.

thus you get that the path $C$ is in fact a closed contour. the integral on the union of the two horizontal lines is equal to the integral on the closed contour $\gamma : -\infty \to +\infty \to +\infty + i\pi \to -\infty +i \pi \to -\infty$

$$\kappa(s) (1-2^{-s/2}) (1-e^{ i \pi s}) = \oint_\gamma e^{su} ( F(u) - F(u + \sqrt{2})) du$$

and you get that $F(u) - F(u + \sqrt{2})$ cannot be an entire function, because if it was, $\kappa(s) (1-2^{-s/2}) (1-e^{ i \pi s})$ would be the zero everywhere function.

finally, because your function $\Phi(u)$ is related to $F'(u)$ and $F''(u)$, it is not entire as well.

this also means that $\kappa(s) (1-2^{-s/2})(1-2^{(1-s)/2}) (e^{2i \pi (1-s)}-e^{ 2 i \pi s})$ can be expressed with the residue theorem, leaving us with a (generalized) Dirichlet series, being entire and even (with respect to $\Re(s) = 1/2$), but this is another story.

• I believe there's a sign problem in one of the exponents of 2 in $\kappa(s) (1-2^{-s/2})(1-2^{(1-s)/2}) (e^{2i \pi (1-s)}-e^{ 2 i \pi s})$. Commented Nov 25, 2018 at 0:48

Regarding question 1, the $\exp(-\pi n^2\exp(2u))$ term will make the series will converge absolutely, and uniformly on compact subsets, as long as $\exp(2u)>0$, i.e., with $u=x+iy$, as long as $-\pi/4<y<\pi/4$ (or in similar horizontal strips translated vertically by multiples of $\pi$.)

This series arises via an (exponential) change of variable and integration by parts in the expression of $\Xi$ as a Mellin transform of a theta function. So (up to the exponential change of variable) it is a derivative of the theta function. See Titchmarsh for details.

• Is it $\text{Re}(\exp(2u))=\exp(2x)\cos(2y)>0$?
– mike
Commented Oct 17, 2014 at 17:44
• Yes. $\ \ \ \ \$ Commented Oct 17, 2014 at 17:50