# Integral of a floor function

• Aug 28th 2009, 06:27 PM
hubric
Integral of a floor function
Greetings.

A bit of help would be much appreciated on the following integral:

$\int_{0}^{\infty} \lfloor x \rfloor e^{-x} dx$

Edit:

Ah this is just a geometric series: $\int_{n-1}^{n} n e^{-x} dx$
• Aug 28th 2009, 06:37 PM
Krizalid
Note $\int_{j}^{j+1}{e^{-x}\,dx}=(e-1)e^{-j-1},$ so $\int_{0}^{\infty }{\left\lfloor x \right\rfloor e^{-x}\,dx}=\sum\limits_{j=0}^{\infty }{\int_{j}^{j+1}{\left\lfloor x \right\rfloor e^{-x}\,dx}}=(e-1)\sum\limits_{j=0}^{\infty }{je^{-j-1}}.$

Computation of that series will give ya the answer.
• Aug 28th 2009, 06:41 PM
hubric
Good catch it looks like we cross posted.