# Quantized calculus

I have heard about a generalization of the calculus named 'quantized calculus'. In this calculus the derivative is defined as

$$df= [F,f]=Ff-fF$$

Here $F(g(x))= \frac{i}{\pi}\int_{-\infty}^{\infty}dt \frac{g(t)-g(s)}{t-s}$. In any case if this is the 'quantized' derivative , how can one defined a 'quantized integral'? How can one recover the usual definition of derivative from this $\frac{d}{dx}$?

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I can tell $F$ is an operator, but I have no idea what $[F,f]$ means nor what $H$ is. In any case, this smells like physics... – anon Apr 24 '12 at 10:45
$[A,B]$ is the commutator between the 2 operator.. or similar , i think $f$ is the element of an algebra. – Jose Garcia Apr 24 '12 at 10:53
I assume by the letter $H$ you actually meant $F$ then right? Would I be correct in also assuming $f$ is a function and $df$ as an operator? – anon Apr 24 '12 at 10:56
ah of course i meant $F$ instead of H :) ... and $df$ should be an operator .. – Jose Garcia Apr 24 '12 at 11:25
I don't know anything at all about quantized calculus, but most of the top hits Google shows me for the phrase state that the analogue of the classical integral in this setting is the "Dixmier trace" (whatever that is). – Rahul Apr 24 '12 at 13:00