Using the well-known integral expression for the Riemann xi-function (without the first factor $\frac12$):
$$\xi(s)=s(s-1)\int^\infty_1 \left(x^{s/2}+x^{(1-s)/2}\right)\frac{\psi(x)}{x}\,\mathrm dx \tag{1}$$
and the relationship
$$\psi(x) = \frac{1}{\sqrt{x}}\,\psi\left(\frac{1}{x}\right)+\frac{1}{2\,\sqrt{x}}-\frac12$$
with $\displaystyle \psi(z)=\sum_{n=1}^\infty e^{-\pi\,n^2z}$, I derived the following equation:
$$\xi(s) = \displaystyle s \left(s-1 \right) \left( \int_{1/z}^{1}\!{\frac {{\it \psi} \left( x \right) }{x} \left( {x}^{s/2}+{x}^{(1-s)/2} \right) }\,{\rm d}x+\int_{z}^{\infty}\!{\frac {{\it \psi} \left( x \right) }{x} \left( {x}^{s/2}+{x}^{(1-s)/2} \right) }\,{\rm d}x \right)\\ +s \left( {z}^{(1-s)/2}+{z}^{(s-1)/2} \right) + \left( 1-s \right) \left( {z}^{s/2}+{z}^{-s/2} \right) -1 \qquad \qquad (2)$$
which is valid for all $s, z \in \mathbb{C}$ with $\Re(z) > 0$.
For $z = 1$ it correctly reduces to (1) and for $z \rightarrow 0$ or $z \rightarrow \infty$ it ends up (as expected) as undetermined or infinite. I wondered whether specific other values of $z$ exist that could help simplify things further. For instance:
- using the critical line: $\Xi(t)=\xi\left(\frac12 + it\right)$,
- making a variable change in both integrals: $\displaystyle x = \mathrm e^{2u},\quad \mathrm dx = 2\mathrm e^{2u}\,\mathrm du,\quad\int_{1/z}^1 \rightarrow \int_{(-\ln z)/2}^0,\quad \int_z^\infty \rightarrow \int_{(\ln z)/2}^\infty $,
- putting $z = \mathrm e^2$,
and doing some reshuffling, yields
$$\displaystyle \Xi(t)= \left( -4\,{t}^{2}-1 \right)\! \left( \int_{0}^{1}\!{\rm e}^{-u/2}\,\psi( {{\rm e}^{-2u}} ) \cos tu \,{\rm d}u +\int_{1}^{\infty}\!{\rm e}^{u/2}\, \psi({{\rm e}^{2u}})\cos tu \,{\rm d}u \right)\\ +4\,t\sinh\!\tfrac12\, \sin t +2\cosh\! \tfrac12\, \cos t-1 ,$$
ADDED: a single integral is achieved through the variable change $w=1/u$ in the first integral: $$\displaystyle \Xi(t)= \left( -4\,{t}^{2}-1 \right)\! \left( \int_{1}^{\infty}\! 1/u^2 \, {\rm e}^{-1/(2u)}\,\psi( {{\rm e}^{-2/u}} ) \cos t/u + {\rm e}^{u/2}\, \psi({{\rm e}^{2u}})\cos tu \,{\rm d}u \right)\\ +4\,t\sinh\!\tfrac12\, \sin t +2\cosh\! \tfrac12\, \cos t-1 ,$$
and this evaluates remarkably fast.
Q: Could there any other choices for $z$ that would help simplify these integrals?
P.S. (just a fun fact):
Isolating $z^{-s/2}$ from (2) and putting $z=n$, with $n \in \mathbb{N}$, provides a working, yet extremely inefficient way to compute $1/n^{s/2}$.