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In the appendix to ONAG (2nd edition), Conway points that the definition of integration (using Riemann sums as left and right options) gives the "wrong" answer : $\int_0^\omega \exp(t)\thinspace dt=\exp(\omega)$ (instead of $\exp(\omega)-1$). I wonder if this was not due to lack of some options (presumably right ones), and if a better definition could not be sought by using Kurzweil-Henstock integration (but I was not able to concoct one, of course, else I would not ask here). Has the idea already been tried?

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For those of us whose surreal calculus skills are a bit rusty, would it be possible for you to sketch the calculation for the "wrong" answer? –  Joel David Hamkins Apr 24 '12 at 11:56
    
Mmm... I cannot find my copy of ONAG ; the main idea is to use for left and right options Riemann sums of the function to be integrated, integrals of the same function over simpler (i.e. given by options of the bounds) intervals, and integrals of simpler functions on the same interval (working only with positive monotonous functions, if I remember well). –  Feldmann Denis Apr 24 '12 at 16:36
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The calculation is attributed to Kruskal but details are not provided. The appendix does give Norton's definition of the integral; this is rather complicated and takes about half a page to explain. Since Conway's conclusion is that Norton's definition is probably not the "right" one, I'm not sure it would be that helpful to reproduce it here. –  Timothy Chow Apr 24 '12 at 21:00

1 Answer 1

http://www.dm.unipi.it/~fornasiero/phd_thesis/thesis_fornasiero_linearized.pdf

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This thesis says in the conclusion that 'A “natural” definition of integral that gives the right answer for the exponential function ... is the main open problem of this thesis.' So that reference doesn't answer the question. –  Carl Feynman Dec 27 '12 at 21:22

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