Hi(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Definite Integral

Loading...

Similar Threads - Definite Integral | Date |
---|---|

I Integrating scaled and translated indicator function | Nov 20, 2017 |

B Definite integrals with +ve and -ve values | Jun 10, 2017 |

I Taylor expansions and integration. | May 7, 2017 |

I Solving a definite integral by differentiation under the integral | Mar 23, 2017 |

I Question about Complex limits of definite integrals | Jan 30, 2017 |

**Physics Forums - The Fusion of Science and Community**