[532nm]

>> I tried to define the integration operator in terms of e^x. The 1 + x/N needed to be one "infinitesimal" iteration of integration, that adds an extra infinitesimal rectangle to the area. But it didn't seem to work out.

You're close! This can indeed be done properly and is then called the Euler-Maclaurin formula. For this, you define the "shift to the left by n operator" e^(nD) where D is the differentiation operator d/dx.

You then always take the current value of f(x), multiply it by the small shift n to get the first rectangle. Then you shift to the left by n, i.e. to e^(nD)*f(x) = f(x+n), multiply that by the small shift n to get the next rectangle etc.

The book "street-fighting mathematics" [1][pdf] has a very hands-on and playful derivation of this in chapter 6.3.

[1] https://mitpress.mit.edu/books/street-fighting-mathematics

[pdf]https://www.dropbox.com/s/722rlvrwy9l9w73/7728.pdf?dl=1

[man-and-laptop]

Amazing, thanks.