Definite Integral

Hi

I need help doing the following integration:

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

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

$$n(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{2}})$$

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 $$k \epsilon Z$$ such that

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

which means that

$$k-1<k-\frac{1}{\sqrt{2}}\leq x - \frac{1}{\sqrt{2}} < k+1-\frac{1}{\sqrt{2}}$$ and
$$k-1<k-\frac{1}{\sqrt{3}}\leq x - \frac{1}{\sqrt{3}} < k+1-\frac{1}{\sqrt{3}}$$

But this would mean that (#)

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

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

Last edited:

Galileo
$$\lfloor x-\sqrt2 \rfloor=\lfloor x-\sqrt3 \rfloor$$ if $$0\leq x < \sqrt{2}$$ or $$\sqrt{3}\leq x \leq 1$$
What happens when $$\sqrt{2} \leq x < \sqrt{3}$$.