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

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- freckles

- jamiebookeater

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- freckles

trying to figure out the floor function thingy

- freckles

got it that is a long command for the floor function

- freckles

anyways...
\[n \le x

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## More answers

- freckles

trying to think how to integrate the greatest integer function

- 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

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

- 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

actually I guess I don't need integration by parts

- ganeshie8

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

- 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

I mean it does diverge sorry

- freckles

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

- ganeshie8

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

- freckles

did something wrong

- freckles

oops didn't mean to delete that without copying it

- 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

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

- freckles

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

- ganeshie8

i pulle that from ur deleted replies

- 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

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

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

- freckles

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

- freckles

well wolfram showed my sum diverged :p

- 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

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

- freckles

ues from that last line

- xapproachesinfinity

ah ok
i guess ration test can testify divergence

- ganeshie8

we can simply use the limit test

- xapproachesinfinity

oh yes! that limit does not go to 0

- 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

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

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

algebra mistakes

- freckles

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

- 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

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

- 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

Thanks so mmmmmmmmuch

- xapproachesinfinity

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

- xapproachesinfinity

- xapproachesinfinity

wrong tagging lol

- 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

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

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

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

ooh i see it now

- 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

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

- freckles

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

- 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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