The product $\xi^*(s)=\xi(s)\,\xi(-s)$, with $\xi(s)$ the Riemann_xi_function, but ignoring the first factor $\frac12$, possesses some 'beauty' in the sense that it yields:
$$\xi^*(s)=s^2\,(s^2-1)\,\Gamma\left(\frac{s}{2}\right)\,\Gamma\left(-\frac{s}{2}\right)\,\zeta(s)\,\zeta(-s) = \prod_{n=1}^\infty \left(1- \left(\frac{s}{\rho_n}\right)^2 \right) \left(1- \left(\frac{s}{1-\rho_n}\right)^2 \right)$$
where $\rho_n$ is the $n$-th non-trivial zero of $\zeta(s)$.
Obviously $\xi^*(s)=\xi^*(-s)$ and it remains an entire function. Its 'elegance' does come at the cost of introducing a second series of $\rho$s at $-\rho_n$.
I wonder whether there could also exist a 'nice' integral expression for $\xi^*(s)$? (with 'nice', I mean not simply a double integral of the known integrals for $\xi(s)$).
ADDED 1: my first (failed) attempt:
Replace $\Gamma\left(\frac{s}{2}\right)\,\Gamma\left(-\frac{s}{2}\right)$ as follows:
$$\xi^*(s)=s\,(1-s^2)\,\frac{2\pi}{\sin\left(\frac{\pi s}{2}\right)}\,\zeta(s)\,\zeta(-s)$$
Start the construction from the following integral representation:
$$\frac{\pi}{\sin\left(\frac{\pi s}{2}\right)}=\int^\infty_0 \frac{t^{\frac{s}{2}}}{t+t^2}\, dt \qquad 0 < \Re(s) < 2$$
Change variable: $t=n^2\,x, dt = n^2\, dx$:
$$\frac{\pi}{\sin\left(\frac{\pi s}{2}\right)}=\int^\infty_0 n^s x^\frac{s}{2}\,\frac{1}{x+(n\,x)^2}\, dx \qquad 1 < \Re(s) < 2$$
Summation on both sides:
$$\frac{\pi}{\sin\left(\frac{\pi s}{2}\right)}\sum^\infty_{n=1} \frac{1}{n^s}=\int^\infty_0 x^\frac{s}{2}\,\sum^\infty_{n=1}\frac{1}{x+(n\,x)^2}\, dx \qquad 1 < \Re(s) < 2$$
Which simplifies into:
$$\frac{2\pi}{\sin\left(\frac{\pi s}{2}\right)}\zeta(s)= \int^\infty_0 x^{\frac{s}{2}-1}\,\left(\frac{\pi}{\sqrt{x}}\,\coth\left(\frac{\pi}{\sqrt{x}}\right)-1\right)\, dx \qquad 1 < \Re(s) < 2$$
Change variable once more: $x=k^{-2}\,u, dx = k^{-2}\, du$ and sum both sides:
$$\frac{2\pi}{\sin\left(\frac{\pi s}{2}\right)}\zeta(s)\sum^\infty_{k=1} \frac{1}{k^{-s}}= \int^\infty_0 u^{\frac{s}{2}-1}\,\sum^\infty_{k=1}\left(\frac{\pi\,k}{\sqrt{u}}\,\coth\left(\frac{\pi\,k}{\sqrt{u}}\right)-1\right)\, du \qquad \infty$$
This obviously fails and unless I missed a tweak, I do need a new approach...
ADDED 2: my second attempt (successful, however with a double integral):
$$\xi^*(s)=s^2(s^2-1)\int^\infty_1\int^\infty_1 \left(x^\frac{s}{2}+x^{\frac{1-s}{2}}\right)\left(y^{-\frac{s}{2}}+y^{\frac{1+s}{2}}\right)\frac{\psi(x)\,\psi(y)}{x\,y}\, dx dy + \xi(s)+\xi(-s) -1$$
With $\displaystyle \psi(z)=\sum_{n=1}^\infty e^{-\pi\,n^2z}$. I did not manage to simplify this any further.