The Riemann $\xi$ and $\Xi$-functions are respectively defined as:
\begin{align} \xi(s) &= \frac{s\,(s-1)}{2}\, \pi^{-s/2} \,\Gamma\left(\frac{s}{2}\,\right) \zeta(s) \qquad s \in \mathbb{C} \\ \Xi(t) &=\xi\left(\frac12 +it\right) \qquad \qquad \qquad \,\,\,\quad t \in \mathbb{C}\\ \end{align}
The Fourier cosine integral expression for $\Xi(t)$ and its inverse are: \begin{align} \Xi(t) &=\,\,2\,\int_0^\infty \Phi(x)\,\cos(xt)\, \mathrm{d} x \tag{1} \\ \Phi(t) &=\frac{1}{\pi}\,\int_0^\infty \Xi(x)\,\cos(xt)\, \mathrm{d} x \tag{2} \end{align}
where $\Phi$ is the super-exponentially decaying even function: \begin{align} \Phi(u) = \Phi(-u) =2\,\sum_{n=1}^\infty (2\,\pi^2 n^4 e^{9u/2} - 3\,\pi\, n^2 e^{5u/2} ) \exp(-\pi\, n^2 e^{2u} ). \end{align}
Setting $t=0$ in (2) yields:
$$\int_0^\infty \Xi(x)\, \mathrm{d} x = \pi\,\Phi(0) = \pi\left(2\,\theta''(1)+3\,\theta'(1)\right) \tag{3}$$
where $\displaystyle \theta(x) =\sum_{n=-\infty}^\infty \exp(-n^2\pi\,x)$, the well-known 3rd Jacobi theta function.
In 2018, Romik published The Taylor coefficients of the Jacobi theta constant $\theta_3$ and provided us with closed forms for $\theta''(1)$ and higher derivatives. We can now simply derive the (new?) closed form for the integral over the critical line that is unconditional on the RH:
\begin{equation} \boxed{\int_0^\infty \Xi\left(x\right)\mathrm{d}x = \Gamma\left(\frac14\right)\cdot\frac{\Gamma\left(\frac14\right)^8-96\,\pi^4}{256\,\sqrt{2}\,\pi^{15/4}}} \tag{4} \end{equation}
In 2011, Lagarias and Montague published the paper The integral of the Riemann $\xi$-function, in which they label the RHS of (3) as $A_0$ (i.e. not the closed form) and observe:
Question: Does such an arithmetic interpretation of the RHS of (4) exist? Is there any new information about $\Xi(t)$ we could derive from the closed form?
Additionally, the same questions could be asked about this related MO-question: Xi Function on Critical Strip - Mellin Transform, that deals with the slightly amended integral (already known by Titchmarsh):
$$\int_0^\infty \frac{\Xi(x)}{x^2 + \frac{1}{4}} \cos(xt) dx = \frac{1}{2} \pi (e^{\frac{1}{2}t} - 2e^{-\frac{1}{2}t} \psi(e^{-2t})) \tag{5}$$
where $\displaystyle \psi(x)=\sum_{n = 1}^{\infty} e^{-n^2 \pi x}$. This has an even simpler closed form (using $\psi(1) = \left(\theta(1)-1\right)/2)$:
\begin{align} \int_0^\infty \frac{\Xi(x)}{x^2 + \frac{1}{4}} dx &= \pi \left(\frac12 -\psi(1) \right)\\ &= \pi \left(1 - \frac12\,\theta(1)\right)\\ &= \pi \left(1-\frac{\Gamma\left(\frac14\right)}{2\,\sqrt{2}\,\pi^{\frac14}}\right)\tag{6}\\ \end{align}
