1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration of a Floor Function

  1. Mar 25, 2014 #1
    Mod note: Thread moved from the Calculus section to here.
    I am having trouble evaluating the integral:

    ∫floor(ln(floor(1/x)))dx from 1/4 to 1/2

    I do not know where or how to start. I probably need a full explanation.

    Thanks for any help!!!
     
    Last edited by a moderator: Mar 25, 2014
  2. jcsd
  3. Mar 25, 2014 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What values does floor(1/x) have for 1/4 < x < 1/2?
     
  4. Mar 25, 2014 #3
    The values of floor(1/x) go from 2 to 4, then taking logs and then the floor of that would give me 0 to 1, but I still do not understand how to find that area under the curve.
     
  5. Mar 25, 2014 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You don't need to worry about floor(1/x) = 4 because it only has that value for x=1/4, and the value of a function at a single point makes no contribution to the integral.

    For what values of x do you have floor(1/x) = 2? How about floor(1/x) = 3?
     
  6. Mar 25, 2014 #5
    You have x = 1/2 and x = 1/3. Then what?
     
  7. Mar 25, 2014 #6

    Mark44

    Staff: Mentor

    WhatTheYock, I moved your thread to this homework section. In future posts be sure to use the homework template, and include the problem statement and your efforts.
     
  8. Mar 25, 2014 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    As jbunniii said, f(x)= 4 only for x= 1/4 which doesn't affect the integral. For [itex]1/4< x\le 1/3[/itex] [itex]3\le \frac{1}{x}< 4[/itex] so f(x)= 3 and for [itex]1/3< x\le 1/2[/itex] [itex]2\le \frac{1}{x}< 3[/itex] so f(x)= 2.

    So [itex]\int_{1/4}^{1/2} f(x) dx= 3(1/3- 1/4)+ 2(1/2-1/3)[/itex].
     
  9. Mar 25, 2014 #8

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    @Halls - Note that the problem is to integrate ##\text{floor}(\ln(\text{floor}(1/x))) = \text{floor}(\ln(f(x)))## assuming you are defining ##f(x) = \text{floor}(1/x)##. But the OP should easily be able to make the appropriate modifications.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Integration of a Floor Function
  1. Floor function proof (Replies: 3)

Loading...