Integration of greatest integer function

• March 12th 2009, 03:58 AM
champrock
Integration of greatest integer function
• March 12th 2009, 04:05 AM
Moo
Hello,
Quote:
Separate the integral :
$\int_0^2 [x]^n f'(x) ~dx=\int_0^1 [x]^n f'(x) ~dx+\int_1^2 [x]^n f'(x)~dx$
if x is between 0 and 1, then [x]=0
if x is between 1 and 2, then [x]=1

so your integral is actually $\int_1^2 f'(x) ~dx$, if $n>0$

if n=0, then the integral is $\int_0^2 f'(x) ~dx$
• March 12th 2009, 04:16 AM
champrock
but its actually [X]^N whcih is most confusing. (mod raised to Nth power).

so, it cant be seperated that easily.
• March 12th 2009, 04:18 AM
Moo
Quote:

Originally Posted by champrock
but its actually [X]^N whcih is most confusing. (mod raised to Nth power).

so, it cant be seperated that easily.

It is... Because x varies from 0 to 1 and from 1 to 2.

x in [0,1[ => [x]=0 => [x]^n=0^n=0, if n is not equal to 0.
x in [1,2[ => [x]=1 => [x]^n=1^n=1.
• March 12th 2009, 04:29 AM
champrock
hmm.. lets just take an example. assume that n=5 and x=1.5, then 1.5^5 = 7.59..
• March 12th 2009, 07:37 AM
Krizalid
Let $k\ge2,$ be an integer then $\int_{0}^{k}{\left\lfloor x \right\rfloor ^{n}f'(x)\,dx}=\sum\limits_{m=0}^{k-1}{\int_{m}^{m+1}{m^{n}f'(x)}\,dx}=\sum\limits_{m= 1}^{k-1}{m^{n}\big(f(m+1)-f(m)\big)}.$

Hence, for $k=2$ it's $\int_{0}^{2}{\left\lfloor x \right\rfloor ^{n}f'(x)\,dx}=f(2)-f(1),$ which is the value of the integral.
• March 12th 2009, 09:07 AM
Moo
Quote:

Originally Posted by champrock
hmm.. lets just take an example. assume that n=5 and x=1.5, then 1.5^5 = 7.59..

But you're working with the greatest integer function !

It's [x]^n, not [x^n], that's certainly not the same...

what I showed you gives the exact same result as Krizalid. (apart from the case n=0)