# [SOLVED] Integration of floor functions

• May 18th 2010, 11:59 AM
dwsmith
[SOLVED] Integration of floor functions
If $\left \lfloor x \right \rfloor$ denotes the greatest integer not exceeding x, then $\int_{0}^{\infty}\left \lfloor x \right \rfloor e^{-x}dx$.

$\int_{0}^{\infty}\left \lfloor x \right \rfloor e^{-x}dx=\sum_{x=0}^{\infty}\left \lfloor x \right \rfloor e^{-x}$

I know this can be manipulated into a telescoping series but I am stuck here.
• May 18th 2010, 01:06 PM
chisigma
Is...

$\int_{0}^{\infty} \lfloor x \rfloor \cdot e^{-x} dx = \int_{0}^{1} 0\cdot e^{-x} dx + \int_{1}^{2} e^{-x} dx + \int_{2}^{3} 2\cdot e^{-x} dx + \dots + \int_{n}^{n+1} n\cdot e^{-x} dx + \dots =$

$e^{-1} - e^{-2} + 2\cdot e^{-2} -2\cdot e^{-3} + \dots + n\cdot e^{-n} - n\cdot e^{-(n+1)} + \dots = e^{-1} + e^{-2} + \dots + e^{-n} + \dots =$

$= \frac{e^{-1}}{1-e^{-1}} = .581976706869...$

Kind regards

$\chi$ $\sigma$
• May 18th 2010, 01:17 PM
dwsmith
Quote:

Originally Posted by chisigma
Is...
$e^{-1} - e^{-2} + 2\cdot e^{-2} -2\cdot e^{-3} + \dots + n\cdot e^{-n} - n\cdot e^{-(n+1)} + \dots = e^{-1} + e^{-2} + \dots + e^{-n} + \dots =$

$= \frac{e^{-1}}{1-e^{-1}} = .581976706869...$

Kind regards

$\chi$ $\sigma$

I am confused though on how you went from the series to fraction
• May 18th 2010, 01:55 PM
chisigma
The expression $\sum_{n=1}^{\infty} e^{-n}$ is a 'geometric series' and its sum is given by a well known formula...

Kind regards

$\chi$ $\sigma$
• May 18th 2010, 03:15 PM
Krizalid
in short way, it's $\int_{0}^{\infty }{\left\lfloor x \right\rfloor e^{-x}\,dx}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{\left\lfloor x \right\rfloor e^{-x}\,dx}}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{je^{-x}\,dx}},$ so the computation of last integral and the remaining series gives the result.
• May 18th 2010, 03:17 PM
dwsmith
Quote:

Originally Posted by Krizalid
in short way, it's $\int_{0}^{\infty }{\left\lfloor x \right\rfloor e^{-x}\,dx}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{\left\lfloor x \right\rfloor e^{-x}\,dx}}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{je^{-x}\,dx}},$ so the computation of last integral and the remaining series gives the result.

I was hoping to see the integral set up as a series minus a series which turns out to be telescoping since I know it can be done that way.