Integration of a Floor Function

[WhatTheYock]

Mod note: Thread moved from the Calculus section to here.[/color]
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!

 

[jbunniii]

What values does floor(1/x) have for 1/4 < x < 1/2?

 

[WhatTheYock]

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.

 

[jbunniii]

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?

 

[WhatTheYock]

You have x = 1/2 and x = 1/3. Then what?

 

[Mark44]

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.

 

[HallsofIvy]

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

So $\int_{1/4}^{1/2} f(x) dx= 3(1/3- 1/4)+ 2(1/2-1/3)$.

 

[jbunniii]

@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.