I'm often interested in generalizing functions, and one of the things I was thinking about recently was the concept of the Riemann definite integral as a function with some rather specific domains. It made me think and wonder about the generalization of the definite integral to the complex plane in the form of the contour integral, and I got to wondering how far this generalization extended, or could be extended.

The Riemann definite integral is usually expressed as:

$$I=\int_{a}^bf(x)\ \mathrm dx.$$

With a and b and I belonging to the reals and f(x) being some continuous function on the reals.

We could define it as a higher order function F(a,b, f) with domains of the reals for a and b and real valued functions for f and codomain of the reals. Now, we have seen these domains and codomains extended to complex numbers with contour integrals, but what other domains can definite integrals be extended to and still be something that is recognizably a definite integral?

Can they be extended to hypercomplex numbers, lie algebras, or various arbitrary algebraic structures and still be recognizably a 'definite integral?'

  • 1
    $\begingroup$ Yes, using measure theory. $\endgroup$ – Matt Samuel Jan 16 '15 at 0:06
  • $\begingroup$ $f$ doesn't have to be continuous; piecewise continuous is certainly enough and I think it can be relaxed further. In higher-dimensional calculus one replaces $\langle a,b\rangle$ with all sorts of things. For example, the notation $\oint_C$ means to integrate around the closed loop $C$. $\endgroup$ – MJD Jan 16 '15 at 0:54
  • $\begingroup$ Even in non-commutative, non-associative domains or domains with zero divisors such as quaternions, octonions, or sedonions? Or finite groups of various sorts? $\endgroup$ – dezakin Jan 16 '15 at 1:02
  • $\begingroup$ I don't see how you could make it work for a nonassociative operation, because in those contexts you don't even get consistent results from finite sums, never mind infinite sums. But I see no obvious reason why it couldn't be extended to a suitable noncommutative operation. Note that quaternion addition is commutative, so extending the integral to quaternion-valued functions wouldn't have noncommutativity as an obstacle. $\endgroup$ – MJD Jan 16 '15 at 1:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.