Question

Fun integral

Answer 1

\[\LARGE \int\limits_0^\infty \lfloor x \rfloor e^{-x}dx\]

Answer 2

Floor function!

Answer 3

\[\int\limits_0^\infty \lfloor x \rfloor e^{-x}dx = \sum\limits_{k=0}^{\infty} k (e^{-k }- e^{-(k+1)})\]

Answer 4

http://www.wolframalpha.com/input/?i=+%5Csum%5Climits_%7Bk%3D1%7D%5E%7B%5Cinfty%7D+k+%28e%5E%7B-%28k%29+%7D-+e%5E%7B-%28k%2B1%29%7D%29

Answer 5

How did you manage to move it over to that series? I'm guessing the jaggedness has something to do with this, but I don't see how you could figure that one out without wolfram alpha.

Answer 6

just split the integral based on intervals in which the floor function is constant

Answer 7

\[\int\limits_0^{\infty} \lfloor x\rfloor e^{-x}dx = \int\limits_0^{1} 0 e^{-x}dx + \int\limits_1^{2} 1 e^{-x}dx + \int\limits_2^{3} 2e^{-x}dx +\cdots \]

Answer 8

Oooooh Ok thanks that's really fancy I like it. =)

Answer 9

you must be having a more fancier method im sure xD

Answer 10

No, I couldn't figure this out, this was a question on a practice exam for getting into math graduate school and it's timed so after trying integration by parts and trying to approximate it (it's multiple choice) I couldn't get it.

Answer 11

ganeshie you made this problem a whole lot easier to understand lol

Answer 12

check this
http://math.stackexchange.com/questions/1060518/integrate-int-1-infty-lfloor-x-rfloor-e-x-dx?rq=1

Answer 13

http://gyazo.com/56f455c27da00c965c508922b4730f31

Answer 14

Hahaha ok I'll try the geometric series version.