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Definite Integral

  1. Dec 10, 2004 #1

    I need help doing the following integration:


    where n is an integer and [.] denotes the greatest integer function (floor), i.e. [x] = greatest integer less than or equal to x.

    The answer given in the book is


    whereas I am getting zero as the answer. My solution is a bit jerky due to the step marked # below:

    By definition, there exists some [tex]k \epsilon Z[/tex] such that

    [tex]k\leq x < k+1[/tex] (so that [x] = k)

    which means that

    [tex]k-1<k-\frac{1}{\sqrt{2}}\leq x - \frac{1}{\sqrt{2}} < k+1-\frac{1}{\sqrt{2}}[/tex] and
    [tex]k-1<k-\frac{1}{\sqrt{3}}\leq x - \frac{1}{\sqrt{3}} < k+1-\frac{1}{\sqrt{3}}[/tex]

    But this would mean that (#)

    [tex][{x-\frac{1}{\sqrt{2}}] = k-1[/tex]
    [tex][{x-\frac{1}{\sqrt{3}}] = k-1[/tex]

    Making the integrand zero and hence the integral zero as well.

    I am not sure if this reasoning is correct (in particular, the integer parts cannot be greater than k-1 so this step could be wrong but still they can attain no other integral value) so I would be very grateful if someone could guide me here.

    Thanks and cheers
    Last edited: Dec 10, 2004
  2. jcsd
  3. Dec 10, 2004 #2


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    Science Advisor
    Homework Helper

    I suggest another method of approach. Draw the graph. Once you have that, the answer is trivial.

    Notice that
    [tex]\lfloor x-\sqrt2 \rfloor=\lfloor x-\sqrt3 \rfloor[/tex] if [tex]0\leq x < \sqrt{2}[/tex] or [tex]\sqrt{3}\leq x \leq 1[/tex]

    What happens when [tex]\sqrt{2} \leq x < \sqrt{3}[/tex].

    What changes when x increases by 1?
    Last edited: Dec 10, 2004
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