Question

Hey! I want to think about evaluating: \[\int_0^\infty \lfloor x\rfloor e^x dx \text{ where } \lfloor . \rfloor\text{ is greatest integer function }\]

Answer 1

trying to figure out the floor function thingy

Answer 2

got it that is a long command for the floor function

Answer 3

anyways...
\[n \le x<n+1 \text{ then } \lfloor x \rfloor=n \text{ where } n \text{ is integer }\]
now I think I want to do integration by parts...

Answer 4

trying to think how to integrate the greatest integer function

Answer 5

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

Answer 6

\[\sum_{n=1}^{\infty} \int\limits _n^{n+1} n \cdot e^x dx\]

Answer 7

\[\sum_{n =1}^\infty n \int\limits _n^{n+1} e^x dx \\ \sum_{n=1}^\infty n e^x|_n^{n+1} \\ \sum_{n=1}^\infty n[e^{n+1}-e^{n}]\]

Answer 8

actually I guess I don't need integration by parts

Answer 9

thats looks very neat, but it doesn't converge right

Answer 10

yeah wolfram says it doesn't diverge
I thought this was suppose to converge those for some reason
did i make a mistake above

Answer 11

I mean it does diverge sorry

Answer 12

you know what maybe if the exponent on the x was negative it would have converged

Answer 13

yeah exponential overtakes any polynomial, floor(x) is cheaper than a linear polynomial

Answer 14

did something wrong

Answer 15

oops didn't mean to delete that without copying it

Answer 16

\[\sum_{n=1}^{\infty}n(e^{-(n+1)}-e^{-n})\]
whatever I know this converges

I just need to find the sum now
I was thinking this maybe a telescoping series

Answer 17

\[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx \\ =\sum_{n=1}^{\infty} \int\limits_{n}^{n+1} n e^{-x} dx \\ =\sum_{n=1}^{\infty} -n e ^{-x}|_n^{n+1} \\ \\ =\sum_{n=1}^{\infty} -n[e^{-(n+1)}-e^{-n}] \\ \text{ ratio test: } \\ \lim_{n \rightarrow \infty} \frac{-(n+1)[e^{-(n+2)}-e^{-(n+1)]}}{-n [e^{-(n+1)}-e^{-n}]}\] \[\lim_{n \rightarrow \infty}-(1+\frac{1}{n}) \frac{ e^{-(n+2)+(n+1)-e^{-(n+1)+(n+1)}}}{e^{-(n+1)+(n+1)}-e^{-n+(n+1)} } \\ -(1+0) \frac{e^{-1}-e^{0}}{e^{0}-e^{1}} =-1 \frac{e^{-1}-1}{1-e} =-1 \frac{1-e}{e-e^2} =\frac{e-1}{e-e^2} \\ =\frac{e-1}{e(1-e)}=\frac{-1}{e}\]

Answer 18

\[(e^{-2}-e^{-1})+2(e^{-3}-e^{-2})+3(e^{-4}-e^{-3})+4(e^{-5}-e^{-4}) +\cdots \]
not exactly a telescoping series the way it is written at last

Answer 19

i pulle that from ur deleted replies

Answer 20

thanks @ganeshie8
now people can know what problem I'm talking about
I'm trying to evaluate:
\[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx\]

Answer 21

mimicing ur earlier work, \[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx = (e-1)\sum\limits_{n=1}^{\infty} \frac{n}{e^n} \]

Answer 22

so the first floor (x) e^x diverge or not?

Answer 23

i showed it diverge so I think it diverge (unless I made some mistake)

Answer 24

well wolfram showed my sum diverged :p

Answer 25

\(\color{blue}{\text{Originally Posted by}}\) @freckles
\[\sum_{n =1}^\infty n \int\limits _n^{n+1} e^x dx \\ \sum_{n=1}^\infty n e^x|_n^{n+1} \\ \sum_{n=1}^\infty n[e^{n+1}-e^{n}]\]
\(\color{blue}{\text{End of Quote}}\)
from tis result ?

Answer 26

I could have use the ratio test to come up with that know

Answer 27

ues from that last line

Answer 28

ah ok
i guess ration test can testify divergence

Answer 29

we can simply use the limit test

Answer 30

oh yes! that limit does not go to 0

Answer 31

\[\sum\limits_{n=1}^{\infty} \frac{n}{e^{nx}}=\sum\limits_{n=1}^{\infty} \frac{d}{dx}e^{-nx}=\frac{d}{dx}\sum\limits_{n=1}^{\infty} e^{-nx}= \frac{d}{dx}\frac{e^{-x}}{1-e^{-x}}=\cdots\]

Answer 32

and we know \[\frac{d}{dx} \frac{e^{-x}}{1-e^{-x}} \\ \frac{d}{dx} \frac{1}{e^x-1} \\ \frac{d}{dx}(e^x-1)^{-1} \\ -1(e^x-1)^{-2}(e^x) \\ - \frac{e^x}{(e^x-1)^2}\]


but we should evaluate this at x=1 since we had that we really wanted to evaluate
\[\int\limits\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx = (e-1)\sum\limits_{n=1}^{\infty} \frac{n}{e^n} \]
\[(e-1) \frac{-e}{(e-1)^2}=\frac{-e}{e-1}\]
but I think we are to get 1/(e-1)

Answer 33

Oh right
\[\int\limits\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx =-\sum_{n=1}^\infty n[e^{-(n+1)}-e^{-n}]
= \frac{e-1}{e}\sum\limits_{n=1}^{\infty} \frac{n}{e^n}\]

Answer 34

algebra mistakes

Answer 35

thanks @ganeshie8
ok I was sorta failing at the algebra a little too
but I see it now

Answer 36

I can't remember who gave me this problem but @Loser66 this is also on your gre practice exam if you want to look at this.

Answer 37

i remember working this problem differently sometime back here in openstudy but google isn't helping at the moment..

Answer 38

And I bet you have answered so many questions since this that going back through your old questions is horrifying and it also doesn't work to well sometimes. Sometimes it automatically throws me out of the looking through my old question thing. Not sure if you know what I mean.

Answer 39

Thanks so mmmmmmmmuch

Answer 40

truly i could not see how you threw differentiation in there @gabylovesu
how is that sum equal to that differentiation ?

Answer 41

@ganeshie8

Answer 42

wrong tagging lol

Answer 43

Have you tried parameterizing and applying the Laplace transform? I'm thinking something along the lines of
\[I(s)=\int_0^\infty \lfloor x\rfloor e^{-sx}\,dx=\mathcal{L}\{\lfloor x\rfloor\}\]
The transform definitely exists because \(\lfloor x\rfloor\) is piecewise continuous and of exponential order.

Answer 44

\[\color{red}{\sum\limits_{n=1}^{\infty} \frac{n}{e^{nx}}=-\sum\limits_{n=1}^{\infty} \frac{d}{dx}e^{-nx}}=-\frac{d}{dx}\sum\limits_{n=1}^{\infty} e^{-nx}= -\frac{d}{dx}\frac{e^{-x}}{1-e^{-x}}=\cdots\]

I think it would be easier to make sense of it by differentiating \(\large \color{red}{e^{-nx}}\) and seeing that you get back the starting expression...

Answer 45

hey @SithsAndGiggles I haven't thought of that but I would be interested in seeing that way if you wanted to show that way. I honestly I haven't seen any laplace transform action in like 10 years so I kind of forgot all of that. :p

Answer 46

The idea would be to express \(\lfloor x\rfloor e^{-x}\) in terms of the step function
\[\theta(x-c)=\begin{cases}1&\text{for }x\ge c\\0&\text{for }x<c\end{cases}\]

Answer 47

ooh i see it now

Answer 48

\[\lfloor x\rfloor e^{-x}=\sum_{c=1}^\infty ce^{-x}\Bigg(\theta(x-c)-\theta(x-c-1)\Bigg)\]
So
\[\begin{align*}\mathcal{L}\left\{\lfloor x\rfloor\right\}&=\int_0^\infty \lfloor x\rfloor e^{-sx}\,dx\\
&=\int_0^\infty \sum_{c=1}^\infty ce^{-sx}\Bigg(\theta(x-c)-\theta(x-c-1)\Bigg)\,dx\\
&=\sum_{c=1}^\infty c\int_0^\infty e^{-sx}\Bigg(\theta(x-c)-\theta(x-c-1)\Bigg)\,dx\\
&=\sum_{c=1}^\infty c\mathcal{L}\{\theta(x-c)-\theta(x-c-1)\}\\
&=\sum_{c=1}^\infty c\frac{e^{-cs}-e^{-(c+1)s}}{s}
\end{align*}\]
Setting \(s=1\) gives the series @ganeshie8 was working with. Nothing new I suppose :P

Answer 49

does seem similar but still pretty @SithsAndGiggles
anyways this is making me hungry
so peace and thanks for the fun

Answer 50

I will come back later though to analyze more deeply what you said

Answer 51

As for the actual result:
\[\sum_{c=1}^\infty c(e^{-c}-e^{-(c+1)})=\sum_{c=1}^\infty c\left(\frac{1}{e^c}-\frac{1}{e^{c+1}}\right)=\frac{1}{e}+\frac{1}{e^2}+\frac{1}{e^3}+\cdots=\frac{1}{1-\frac{1}{e}}-1\]