Let $Df$ denote the derivative of a function $f(x)$ and $\bigtriangledown f=f(x)-f(x-1)$ be the discrete derivative. Using the Taylor series expansion for $f(x-1)$, we easily get $\bigtriangledown = 1- e^{-D}$ or, by taking the inverses,
$$ \frac{1}{\bigtriangledown} = \frac{1}{1-e^{-D}} =
\frac{1}{D}\cdot \frac{D}{1-e^{-D}}=
\frac{1}{D} + \frac12+
\sum_{k=1}^{\infty} B_{2k}\frac{D^{2k-1}}{(2k)!}
,$$
where $B_{2k}$ are Bernoulli numbers.
(Edit: I corrected the signs to adhere to the most common conventions.)
Here, $(1/D)g$ is the opposite to the derivative, i.e. the integral; adding the limits this becomes a definite integral $\int_0^n g(x)dx$. And $(1/\bigtriangledown)g$ is the opposite to the discrete derivative, i.e. the sum $\sum_{x=1}^n g(x)$. So the above formula, known as Euler-Maclaurin formula, allows one, sometimes, to compute the discrete sum by using the definite integral and some error terms.
Usually, there is a nontrivial remainder in this formula. For example, for $g(x)=1/x$, the remainder is Euler's constant $\gamma\simeq 0.57$. Estimating the remainder and analyzing the convergence of the power series is a long story, which is explained for example in the nice book "Concrete Mathematics" by Graham-Knuth-Patashnik. But the power series becomes finite with zero remainder if $g(x)$ is a polynomial. OK, so far I am just reminding elementary combinatorics.
Now, for my question. In the (Hirzebruch/Grothendieck)-Riemann-Roch formula one of the main ingredients is the Todd class which is defined as the product, going over Chern roots $\alpha$, of the expression $\frac{\alpha}{1-e^{-\alpha}}$. This looks so similar to the above, and so suggestive (especially because in the Hirzebruch's version
$$\chi(X,F) = h^0(F)-h^1(F)+\dots = \int_X ch(F) Td(T_X)$$
there is also an "integral", at least in the notation) that it makes me wonder: is there a connection?
The obvious case to try (which I did) is the case when $X=\mathbb P^n$ and $F=\mathcal O(d)$. But the usual proof in that case is a residue computation which, to my eye, does not look anything like Euler-Maclaurin formula.
But is there really a connection?
An edit after many answers: Although the connection with Khovanskii-Pukhlikov's paper and the consequent work, pointed out by Dmitri and others, is undeniable, it is still not obvious to me how the usual Riemann-Roch for $X=\mathbb P^n$ and $F=\mathcal O(d)$ follows from them. It appears that one has to prove the following nontrivial
Identity: The coefficient of $x^n$ in $Td(x)^{n+1}e^{dx}$ equals
$$\frac{1}{n!}
Td(\partial /\partial h_0) \dots Td(\partial /\partial h_n)
(d+h_0+\dots + h_n)^n |_{h_0=\dots h_n=0}$$
A complete answer to my question would include a proof of this identity or a reference to where this is shown. (I did not find it in the cited papers.) I removed the acceptance to encourage a more complete explanation.
Best Answer
As far as I understand this connection was observed (and generalised) by Khovanskii and Puhlikov in the article
A. G. Khovanskii and A. V. Pukhlikov, A Riemann-Roch theorem for integrals and sums of quasipolynomials over virtual polytopes, Algebra and Analysis 4 (1992), 188–216, translation in St. Petersburg Math. J. (1993), no. 4, 789–812.
This is related to toric geometry, for which some really well written introduction articles are contained on the page of David Cox http://www3.amherst.edu/~dacox/
Since 1992 many people wrote on this subject, for example
EXACT EULER MACLAURIN FORMULAS FOR SIMPLE LATTICE POLYTOPES
http://arxiv.org/PS_cache/math/pdf/0507/0507572v2.pdf
Or Riemann sums over polytopes http://arxiv.org/PS_cache/math/pdf/0608/0608171v1.pdf