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The Riemann $\Xi$ function, defined as $$ \Xi(z) \equiv -\frac{1}{2}\left(z^2+\frac{1}{4}\right)\pi^{\frac{1}{4}+i\frac{z}{2}}\Gamma\left(\frac{1}{4}+i\frac{z}{2}\right)\zeta\left(\frac{1}{2}+iz\right) $$ has a number of nice properties. It's an entire function, unlike the $\Gamma$ and $\zeta$ functions. Its reflection formula $\Xi(-z) = \Xi(z)$ is particularly easy to remember. The Riemann hypothesis for $\Xi(z)$ is also much simpler: all zeros of $\Xi(z)$ are real.

On the other hand, that formula in its definition is a pretty ugly one, and makes it obvious it's pretty much just the zeta function shifted, rotated, and scaled. Is there a nicer representation for it, possibly in terms of an integral or other special functions?

eyeballfrog
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  • No. The other representations come from the other representations of $\zeta(s)$ in particular those in term of the Mellin transform of $\theta(x)-1 = 2\sum_{n=1}^\infty e^{-\pi n^2 x}$ – reuns Jul 12 '19 at 18:39
  • One answer is the De Bruijn Newman approach to RH which has become current with recent results: https://terrytao.wordpress.com/2018/01/19/the-de-bruijn-newman-constant-is-non-negativ/ – Conrad Jul 12 '19 at 18:41
  • @Conrad Do you have an idea why the roots $x_j(t)$ would move following $\partial t x_j(t)= 2\sum{k \ne j} 1/(x_k(t)-x_j(t))$ ? $\xi_{T-t}(s)$ is the (vertical lines) convolution of $\xi_{T}(s) = \int_{-\infty }^\infty e^{T u^2} \Phi(u)e^{su}du$ with a gaussian $ \frac{\sqrt{\pi}}{t} e^{s^2/ (4t)}$ so for $t$ very small an off-line simple zero of $\xi_T(s)$ will become an off-line simple zero of $\xi_{T-t}(s)$ – reuns Jul 12 '19 at 21:02
  • In https://arxiv.org/pdf/1801.05914.pdf – reuns Jul 12 '19 at 21:08
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    @reuns by general Hadamard stuff we know $H_t(x)=H_t(0)\Pi{(1-\frac{x^2}{x_j(t)^2})}, \Sigma{|x_j(t)|^{-2}} < \infty, 0<|x_1(t)| \le| x_2(t)|...$; if $t> \Lambda$ so we KNOW the zeros are simple and real, we get that $x_j(t)$ is a well defined differentiable function and $x'_j(t)=\frac{H''_t(x_j(t))}{H'_t(x_j(t))}$ by differentiating $H_t(x_j(t))=0$ and using the form of $H_t$; then it is just a general analytic manipulation of the form $f(z)=(z-w)q(z), q(w) \ne 0$ imply: $f'(w) \ne 0, \frac{f''(w)}{f'(w)}=2\frac{q'(w)}{q(w)}$ applied to the Hadamrd product and the above equation of $x'_j(t)$ – Conrad Jul 12 '19 at 21:45
  • In the above the derivative is of course $\partial_t$ for $x_j$ but the usual complex derivative for $H', H''$ – Conrad Jul 12 '19 at 21:56
  • For $\tau$ very small using the convolution formula saying $H_{t-\tau}(x)$ depends only on $H_t(x\pm \epsilon)$ from $H_t'(x_j(t)),H_t''(x_j(t))$ I can estimate $H_{t-\tau}(x_j(t))$ and $H_{t-\tau}'(x_j(t))$ to estimate the location of $x_j(t-\tau) \approx x_t(t) - \frac{H_{t-\tau}(x_j(t))}{H_{t-\tau}'(x_j(t))}$ telling me $x_j'(t)$ – reuns Jul 13 '19 at 22:41

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