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I'm asking this because in many applications where there is a discrete case of some calculation (that includes a summation) and another continuous case, in the continuous one we simply swap the summation sign for an integral.

Note: for the sake of disambiguation, I'm not referring to the Riemann Sum.

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    $\begingroup$ Yes and no. I believe in measure theory, you can argue that summation IS integration when you're in a discrete space, but I know nothing about measure theory so don't hold me on that. $\endgroup$ Apr 16, 2020 at 18:45
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    $\begingroup$ you can check section 2.6 of this book for the connection Graham, R. L., Knuth, D. E., Patashnik, O., & Liu, S. (1994). Concrete mathematics: a foundation for computer science. $\endgroup$
    – Daniel S.
    Apr 16, 2020 at 18:48

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Yes. In fact, if we consider Lebesgue integration, we can view sums as a specific example of an integral (namely, integration wrt counting measure over some countable set).

In probability theory, this also means that we can often prove statements for both continuous and discrete random variables at the same time, by viewing the sums we find in the discrete case as integrals as well. This makes the theory a lot more elegant compared to having to repeat the proofs and definitions for the continuous case and discrete case separately.

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    $\begingroup$ This may be a pointless nitpick, but I think that your answer argues not that integration is a kind of "continuous summation," but rather that summation is a kind of "discrete integration". $\endgroup$
    – Xander Henderson
    Apr 17, 2020 at 5:04
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    $\begingroup$ I agree with Xander - it's usually not complicated to go from an integral to a discrete sum, but OP is asking about the much more difficult question of when is it okay to turn a sum into an integral $\endgroup$
    – GSofer
    Apr 17, 2020 at 6:02
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The short answer is - it depends. There are lots of cases where this is a healthy way of viewing the integral, and it is indeed the initial motivation behind the idea of Riemann sums. But in some cases, you need to be more careful while taking "the continuum limit".

This is especially important in problems in statistical mechanics and fluid dynamics, where you sometimes approximate discrete systems as continuous ones. For instance, there is a very standard derivation in statistical mechanics that shows the existence of a phenomenon known as Bose-Einstein condensate, in which if you take the continuum limit by turning a sum into an integral, you get an incorrect result.

But yeah, generally, in many cases it is fine and even a good way to view the idea of an integral (just wanted to also point out that it might be inaccurate at times, and should be done with caution and with proper understanding of the problem).

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  • $\begingroup$ What are the reasons that make the integral lose this interpretation of continuous summation (for example such as in the instance you gave) ? $\endgroup$
    – Metrician
    Apr 16, 2020 at 22:53
  • $\begingroup$ Bose-Einstein condensate is a very special (and unusual) state of matter where there is a macroscopic number of particles which populate the ground state of the system. Since not only the states of the quantum system are discrete, but the ground state has a very 'different' nature than the rest of the states, then while taking a certain sum over all possible states of the system,you cannot approximate it via an integral, since it 'loses' the special role of the ground state, and you end up 'missing' the macroscopic population of the ground state. $\endgroup$
    – GSofer
    Apr 17, 2020 at 5:44
  • $\begingroup$ The main issue in this case is not necessarily the fact that you approximate quantum states as continuous ones, but the fact that taking the continuum limit makes you lose the ability to see 'specific' states of the system - when you have a continuum, you can't see points, only intervals, and so when a physical phenomenon arises from one specific state, you will miss it completely $\endgroup$
    – GSofer
    Apr 17, 2020 at 5:47
  • $\begingroup$ I would say this just means that an integral is not the appropriate tool for the job in that case, and the failure of the "continuous summation" metaphor to apply is simply saying "this should not be described by an integral" because an integral is continuous summation and this situation is not and we cannot even treat it as approximately such under the parameters given. $\endgroup$ Apr 17, 2020 at 7:10
  • $\begingroup$ OP asked if the integral can be viewed as continuous summation, since while performing calculations of both discrete and continuous systems, he sometimes simply swaps summation by integration. This is an example of when this cannot be done (and there are plenty more others). Just as you said, not all discrete sums correspond (nicely) to integrals, and so integration is not always the discrete analog of summation, since it might give inaccurate results while taking the continuous limit. $\endgroup$
    – GSofer
    Apr 17, 2020 at 7:52

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