freckles
  • freckles
Hey! I want to think about evaluating: \[\int_0^\infty \lfloor x\rfloor e^x dx \text{ where } \lfloor . \rfloor\text{ is greatest integer function }\]
Mathematics
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freckles
  • freckles
Hey! I want to think about evaluating: \[\int_0^\infty \lfloor x\rfloor e^x dx \text{ where } \lfloor . \rfloor\text{ is greatest integer function }\]
Mathematics
jamiebookeater
  • jamiebookeater
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freckles
  • freckles
trying to figure out the floor function thingy
freckles
  • freckles
got it that is a long command for the floor function
freckles
  • freckles
anyways... \[n \le x

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freckles
  • freckles
trying to think how to integrate the greatest integer function
freckles
  • freckles
\[\int\limits_0^\infty \lfloor x \rfloor e^x dx \\ \int\limits_0^1 0 \cdot e^x dx+\int\limits_1^2 1 \cdot e^x + \int\limits_2^3 2 \cdot e^x + \cdots + \int\limits_n^{n+1} n \cdot e^{x} dx +\cdots \]
freckles
  • freckles
\[\sum_{n=1}^{\infty} \int\limits _n^{n+1} n \cdot e^x dx\]
freckles
  • freckles
\[\sum_{n =1}^\infty n \int\limits _n^{n+1} e^x dx \\ \sum_{n=1}^\infty n e^x|_n^{n+1} \\ \sum_{n=1}^\infty n[e^{n+1}-e^{n}]\]
freckles
  • freckles
actually I guess I don't need integration by parts
ganeshie8
  • ganeshie8
thats looks very neat, but it doesn't converge right
freckles
  • freckles
yeah wolfram says it doesn't diverge I thought this was suppose to converge those for some reason did i make a mistake above
freckles
  • freckles
I mean it does diverge sorry
freckles
  • freckles
you know what maybe if the exponent on the x was negative it would have converged
ganeshie8
  • ganeshie8
yeah exponential overtakes any polynomial, floor(x) is cheaper than a linear polynomial
freckles
  • freckles
did something wrong
freckles
  • freckles
oops didn't mean to delete that without copying it
freckles
  • freckles
\[\sum_{n=1}^{\infty}n(e^{-(n+1)}-e^{-n})\] whatever I know this converges I just need to find the sum now I was thinking this maybe a telescoping series
ganeshie8
  • ganeshie8
\[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx \\ =\sum_{n=1}^{\infty} \int\limits_{n}^{n+1} n e^{-x} dx \\ =\sum_{n=1}^{\infty} -n e ^{-x}|_n^{n+1} \\ \\ =\sum_{n=1}^{\infty} -n[e^{-(n+1)}-e^{-n}] \\ \text{ ratio test: } \\ \lim_{n \rightarrow \infty} \frac{-(n+1)[e^{-(n+2)}-e^{-(n+1)]}}{-n [e^{-(n+1)}-e^{-n}]}\] \[\lim_{n \rightarrow \infty}-(1+\frac{1}{n}) \frac{ e^{-(n+2)+(n+1)-e^{-(n+1)+(n+1)}}}{e^{-(n+1)+(n+1)}-e^{-n+(n+1)} } \\ -(1+0) \frac{e^{-1}-e^{0}}{e^{0}-e^{1}} =-1 \frac{e^{-1}-1}{1-e} =-1 \frac{1-e}{e-e^2} =\frac{e-1}{e-e^2} \\ =\frac{e-1}{e(1-e)}=\frac{-1}{e}\]
freckles
  • freckles
\[(e^{-2}-e^{-1})+2(e^{-3}-e^{-2})+3(e^{-4}-e^{-3})+4(e^{-5}-e^{-4}) +\cdots \] not exactly a telescoping series the way it is written at last
ganeshie8
  • ganeshie8
i pulle that from ur deleted replies
freckles
  • freckles
thanks @ganeshie8 now people can know what problem I'm talking about I'm trying to evaluate: \[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx\]
ganeshie8
  • ganeshie8
mimicing ur earlier work, \[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx = (e-1)\sum\limits_{n=1}^{\infty} \frac{n}{e^n} \]
xapproachesinfinity
  • xapproachesinfinity
so the first floor (x) e^x diverge or not?
freckles
  • freckles
i showed it diverge so I think it diverge (unless I made some mistake)
freckles
  • freckles
well wolfram showed my sum diverged :p
xapproachesinfinity
  • xapproachesinfinity
\(\color{blue}{\text{Originally Posted by}}\) @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 ?
freckles
  • freckles
I could have use the ratio test to come up with that know
freckles
  • freckles
ues from that last line
xapproachesinfinity
  • xapproachesinfinity
ah ok i guess ration test can testify divergence
ganeshie8
  • ganeshie8
we can simply use the limit test
xapproachesinfinity
  • xapproachesinfinity
oh yes! that limit does not go to 0
ganeshie8
  • ganeshie8
\[\sum\limits_{n=1}^{\infty} \frac{n}{e^{nx}}=\sum\limits_{n=1}^{\infty} \frac{d}{dx}e^{-nx}=\frac{d}{dx}\sum\limits_{n=1}^{\infty} e^{-nx}= \frac{d}{dx}\frac{e^{-x}}{1-e^{-x}}=\cdots\]
freckles
  • freckles
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)
ganeshie8
  • ganeshie8
Oh right \[\int\limits\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx =-\sum_{n=1}^\infty n[e^{-(n+1)}-e^{-n}] = \frac{e-1}{e}\sum\limits_{n=1}^{\infty} \frac{n}{e^n}\]
ganeshie8
  • ganeshie8
algebra mistakes
freckles
  • freckles
thanks @ganeshie8 ok I was sorta failing at the algebra a little too but I see it now
freckles
  • freckles
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.
ganeshie8
  • ganeshie8
i remember working this problem differently sometime back here in openstudy but google isn't helping at the moment..
freckles
  • freckles
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.
Loser66
  • Loser66
Thanks so mmmmmmmmuch
xapproachesinfinity
  • xapproachesinfinity
truly i could not see how you threw differentiation in there @gabylovesu how is that sum equal to that differentiation ?
xapproachesinfinity
  • xapproachesinfinity
xapproachesinfinity
  • xapproachesinfinity
wrong tagging lol
anonymous
  • anonymous
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.
ganeshie8
  • ganeshie8
\[\color{red}{\sum\limits_{n=1}^{\infty} \frac{n}{e^{nx}}=-\sum\limits_{n=1}^{\infty} \frac{d}{dx}e^{-nx}}=-\frac{d}{dx}\sum\limits_{n=1}^{\infty} e^{-nx}= -\frac{d}{dx}\frac{e^{-x}}{1-e^{-x}}=\cdots\] 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...
freckles
  • freckles
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
anonymous
  • anonymous
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
xapproachesinfinity
  • xapproachesinfinity
ooh i see it now
anonymous
  • anonymous
\[\lfloor x\rfloor e^{-x}=\sum_{c=1}^\infty ce^{-x}\Bigg(\theta(x-c)-\theta(x-c-1)\Bigg)\] 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
freckles
  • freckles
does seem similar but still pretty @SithsAndGiggles anyways this is making me hungry so peace and thanks for the fun
freckles
  • freckles
I will come back later though to analyze more deeply what you said
anonymous
  • anonymous
As for the actual result: \[\sum_{c=1}^\infty c(e^{-c}-e^{-(c+1)})=\sum_{c=1}^\infty c\left(\frac{1}{e^c}-\frac{1}{e^{c+1}}\right)=\frac{1}{e}+\frac{1}{e^2}+\frac{1}{e^3}+\cdots=\frac{1}{1-\frac{1}{e}}-1\]

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