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April 30th, 2010, 11:10 PM   #1
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Riemann/Darboux Integral

Hello to everyone.
I have a question. Is the Riemann Integral really equivalent to Darboux Integral?.
For example, consider the function f(x)=sin(x)/x. I want to know if it is integrable between -1 and 1. The problem arises at x=0.
Now, this function is clearly Darboux integrable, since you solve the problem of the non-existance of f(0) with the supremum operator. But I don't think it is Riemann integrable, since you can not evaluate f(0), and the Riemann definition clearly states that the sum must exist for ANY election of evaluation points. If you defined f(0) as 1, you would not be solving the problem, because you would be integrating a different function,not f(x), since you are changing the domain of the function.
So, is it possible to say that the Riemann and Darboux integrals are equivalent iff the function is well defined for all points in the interval?.
Please correct me if I am wrong.
I will deeply appreciate any replys.
DavidDante is offline  
May 1st, 2010, 12:46 PM   #2
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Re: Riemann/Darboux Integral

I would call your question a quibble, since you want to leave sinx/x undefined at x=0, rather than accepting any value at all. Whatever you define it to be, the resultant function will be Riemann integrable, since the x interval containing x=0 can be made as small as you want, so its contribution -> 0, no matter what the (finite) value assigned at that point.
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