Is there something between summation and integration? Let's take a general function $f(x)$, we can do a summation like:
$$\sum_{k=m}^n f(k)$$
And we can do an integration like:
$$\int_a^bf(k)dk$$
The basic difference between the two operation is that the first concerns only integer values of $k$ while the second is about real values of the function in an interval. So my question is: is there a similar operation that concerns rational values ? Something "between" summation and integration.
 A: One might say: Between sums (finitely many summands) and integrals (uncountably many "summands"), we have series (countably many summands). 
A: The short answer is yes, but it is not a very easy concept. The answer lies with the definition of the integral and measures. The idea of a measure is to assign a size to subsets of the reals, and use this idea of size to perform integration (among other things). The first thing to note is that there are different ways to assign a measure to subsets of the reals. Some measures behave the way that you would expect them to in that the size of the interval $(a,b)$ is $b-a$, regardless of what $a$ and $b$ are. Other measures behave differently. For example, you could say that the measure of a set is the number of integers contained therein. Under this definition of a measure, summation and integration become the same thing. We could also define a measure which heeds only rational numbers and perform integration with respect to that measure. 
In summary, integration is an abstract concept. The standard integral taught in introductory calculus is just one type of integral, and the type of integral you describe is just a different type (as is summation).
For more information on this subject see here for information on measures, and here for an explanation of how integration is defined abstractly, using measures.
A: Just to follow on from something inspired by john-mangual's answer:
Firstly one might know that the rationals $\mathbb{Q}$ are countable and thus can be enumerated by the positive integers $\mathbb{N}$. Let's pick an enumeration  $\mathbb{Q}=\{q_1,q_2,q_3,\ldots\}$. Now let $f:\Bbb R \rightarrow \Bbb R$ such that it is defined on $\mathbb{Q}$ (or restrict to a subset if necessary), then we can sum over the rationals by its enumeration, namely $$\sum\limits_{i=1}^\infty f(q_i).$$
Now unless the above summation converges absolutely, as in unless $$\sum\limits_{i=1}^\infty |f(q_i)| <\infty,$$
then the value of this series may depend on the enumeration of $\mathbb{Q}$, and even worse due to a theorem of Riemann for what are called conditionally convergent series, if you rearrange the terms you can make the series converge to whatever you like! 
However if it does converge absolutely then the sum is independent of how we choose to order the rationals, and we can sum in any order.
Maybe this will give you another insight into something else you can do, and since I don't really know any measure theory I can't help on that front!
A: Your examples can be viewd as particular cases of a general theory of integration, namely, the Lebesgue integration. In fact:


*

*If $X=\{m,m+1,...,n\}$, $\mu$ is the counting measure and $f$ is positive, then $$\int_X f\ d\mu=\sum_{k=m}^nf(k).$$

*If $X=[a,b]$, $\mu$ is the Lebesgue measure and $f$ is continuous, then
$$\int_X f\ d\mu=\int_a^bf(k)\ dk,$$
where the integral in the right side is the Riemann integral.
From this point of view, the answer to your question is yes and the desired "similar operation" will also be a particular case of the Lebesgue integration. Explicitly:


*

*If $X=\mathbb{Q}$, $\mu$ is some suitable measure and $f$ is measurable, then
$$\int_X f\ d\mu$$
can be viewed as "something between summation and (Riemann) integration that concerns rational values".


Example: If $\mu$ is the Lebesgue measure and $f$ is Lebesgue measurable, then
$$\int_\mathbb{Q} f\ d\mu=0.$$
A: Not sure that my answer is totally relevant... Ler's have a try however.
Both cases that you provide can be interpreted as distribution. The integral is a distribution with a "constant density".
$$\sum_{k=m}^n f(k)$$ is also a distribution with Dirac comb for density.
You can imagine a distribution with "masses" only at the rational numbers which will correspond to your request.
A: Certainly for the exponential function you can define fractional derivatives or negative dervatives
$$ \frac{d}{dx^k} [e^{ax}] =  a^k \; e^{ax}$$
where $k \in \mathbb{R}$.  This can be extended to Fourier series so we can define such pseuo-differential operators for all of $L^2 ([0,1])$.
Wikipedia explains such a fractional deriviative behaves nicely with respect to the Laplace transform as well.
$$  \frac{d^k}{dx^k} \int_0^\infty e^{-xt} f(t) \, dt
 = \int_0^\infty t^ke^{-xt} f(t) \, dt$$
with some additional fine print about passing under the integral sign.
Even polynomials seem to behave nicely under this definition
$$ \frac{d^\ell}{dx^\ell} \frac{x^k}{k!} = \frac{x^{k-\ell}}{(k-\ell)!}$$
where $l! = \Gamma(l+1)$ for $l \in \mathbb{R}$ is the Gamma function.

To answer the question you asked, the rationals are countable but dense in the real line $\overline{\mathbb{Q}} = \mathbb{R}$ so there should not be much difference between summing over rationals and integrating unless your function is very chaotic.  So I am arguing
$$ \sum_{x \in \mathbb{Q} \cap [0,1]} \approx \int_0^1 dx\, $$
However, there are many functions on the rational numbers were nearby inputs take radically different values (as appear in the theory of modular forms).
