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Hi

I need help doing the following integration:

[tex]\int_{x=0}^{x=n}[{x-\frac{1}{\sqrt{2}}]-[{x-\frac{1}{\sqrt{3}}]dx[/tex]

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

[tex]n(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{2}})[/tex]

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

Vivek

I need help doing the following integration:

[tex]\int_{x=0}^{x=n}[{x-\frac{1}{\sqrt{2}}]-[{x-\frac{1}{\sqrt{3}}]dx[/tex]

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

[tex]n(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{2}})[/tex]

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

Vivek

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