I initially thought that symbol ⨋ is some sort of joke, but apparently it is used, and it made it's way to Unicode in 2002. The only reference to its meaning or usage I found is its site on wiktionary, which references "The principles of modern thermodynamics" by Deffner and Campbell:
The distribution of work values is then given by averaging over an ensemble of realizations of the same process, $ P(W) = \left< \delta(W - W[\left| m\right>; \left| n \right>] )\right>$ which can be rewritten as $ P(W) = \operatorname{⨋}_{m, n}\delta(W - W[\left| m\right>; \left| n \right>] )p(\left|m\right> \to \left|n\right>)$. In the latter equation the symbol ⨋ denotes that we have to sum over the discrete part of the eigenvalue spectrum and integrate over the continuous part.
I admit that I have no Idea what author is talking about, but it seems that the "summation over the discrete part and integration over the continuous part" could be achieved by integrating with respect to adequate measure. Since integral itself can be viewed as generalization of summation, why bother with this ⨋ symbol?
I think it could be some sort of notation convention used by physicists to link visuals to the method of computation involved, but that's just my blind guess.
Does this symbol have formal (precise) meaning? Where is it used?
I have seen this in physics as a way to deal with distributions containing $\delta$-terms. This often occurs when a physicist extends a discrete situation by analogy to a continuous situation.
For example if we have $k$ point particles situated at $x_1, \dots, x_k \in \mathbb{R}^d$ with respective masses $m_1, \dots, m_k \in \mathbb{R}_{> 0}$, one can define their center of mass as
$$ X = \sum_{j=1}^k \frac{m_j}{\sum_{i=1}^k m_i} x_i. $$
One can extend this to the continuous situation as follows: Formalise the collection of point particles as a function $m: \mathbb{R}^d \to \mathbb{R}_{>0}$ by defining $ m(x) = \sum_{j=1}^d \delta_{x, x_j} m_j $. The function is interpreted as returning the mass $m(x)$ at position $x$. The centre of mass of $m$ then becomes
$$ X = \sum_{x \in \mathbb{R}^d} x \frac{m(x)}{\sum_{x \in \mathbb{R}^d} m(x)}. $$
If one wants to model a continuous mass distribution, one could do this by looking at a function $\rho: \mathbb{R}^d \to \mathbb{R}_{> 0}$, interpreting the integral
$$ \int_{\Omega} \rho(x) \mathrm{d}(x) $$
as the mass contained in the region $\Omega \subseteq \mathbb{R}^d$. By the analogy ,,integrals are just infinite sums'', one finds the generalised center of mass as
$$ X = \int_{\Omega} x \frac{\rho(x)}{\int_{\Omega} \rho(y) \mathrm{d}y} \mathrm{d}x. $$
Now given a general function $f: \mathbb{R}^d \to \mathbb{R}_{>0}$ one has to choose between two options:
The physicist then writes
$$ X = \int\kern-1.5em\sum_{} x \frac{f(x)}{\int\kern-1em\sum_{} f(x)} $$
to indicate that the sensible choice has to be made.
A more mathematically minded person would probably just formalise the mass as a measure or more generally a distribution on $\mathbb{R}^d$ and then could just write an integral regardless. But well...