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I'd like some help with the following integral:

$$\int_0^\infty \lfloor x \rfloor e^{-x}\mathrm dx .$$

Thanks.

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    HINT What can you say about $\lfloor x \rfloor$ if $n \leq x < n+1$ where $n$ is a nonnegative integer?2011-09-13
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    probably a silly question: does not this function have 'kinks' and therefore not integrable?2011-09-13
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    @sigma: it's best interpreted as the series in lhf's answer.2011-09-14
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    When integrating piecewise-defined functions, integrate piecewisely.2015-01-14
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    @fonini: but integral is only finitely-additive.2015-01-14
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    Integrate from 0 up to $t$ and let $t \to \infty$ then.2015-01-14
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    The usual definition of $\int_0^\infty$ is $\lim_{x\to\infty}\int_0^x$. For each $x$ finite, you can write $\int_0^x=\int_0^1+\int_1^2+\cdots+\int_{\lfloor x\rfloor -1}^{\lfloor x\rfloor}+\int_{\lfloor x\rfloor}^x$. Then let $x\to\infty$.2015-01-14

3 Answers 3

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No need for any explicit summing. (Note that we have $\lfloor x+n \rfloor = \lfloor x \rfloor +n$ for any integer $n$.)

Let $I = \int_0^\infty \lfloor x \rfloor e^{-x} dx $. Letting $t=x+1$, we obtain $I = \int_1^\infty (\lfloor t \rfloor -1)e e^{-t} dt = e\int_1^\infty \lfloor t \rfloor e^{-t} dt - e \int_1^\infty e^{-t} dt = e (I-e^{-1})$.

Solving gives $I={1 \over e-1}$.

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    You are using both $x$ and $t$ in the integrals. Once you have made the substitution of $t = x+1$, I think you should use it everywhere. Your argument may be correct, but I find your presentation confusing.2015-01-14
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    @martycohen: Thanks for catching that - it was a cut and paste induced error.2015-01-14
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    I got a 'necromancer' badge for this even though I answered it 3 hours ago!2015-01-14
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    Now it's clear to me. It's nice that the integral from 0 to 1 is zero, so the integral from 0 to infinity is the same as the integral from 1 to infinity. I wonder if this could help for other functions than exp(-x).2015-01-15
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    @martycohen: I don't know, $\exp$ has the nice property that $e^{-(t-1)} = e e^{-t}$ which (along with your $0\to 1$ observation) makes it work.2015-01-15
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$$\begin{align}\int_0^{\infty} dx \, \lfloor x \rfloor \, e^{-x} &= \sum_{k=0}^{\infty} k \int_k^{k+1} dx \, e^{-x} \\ &= \sum_{k=0}^{\infty} k \, \left (e^{-k}-e^{-(k+1)} \right ) \\ &= \left ( 1-e^{-1} \right )\sum_{k=0}^{\infty} k \, e^{-k} \\ &= \left ( 1-e^{-1} \right ) \frac{e^{-1}}{(1-e^{-1})^2} \\ &= \frac1{e-1} \end{align}$$

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    Why is the $dx$ not placed at the end of the integral expression?2015-01-14
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    @dragon this is just a matter of notation. Some people place it just after the integral sign. Just different from what you might be used to, not "wrong" at all.2015-01-14
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    I realize there is always some "abuse of notation" used on a regular basis, but I didn't think this one was acceptable. In fact, this one seems misleading, because at first I read it as $$\left. \int_0^\infty \, dx \lfloor x \rfloor \, e^{-x} = x \right\vert_0^\infty \lfloor x \rfloor \, e^{-x} = \infty \cdot \lfloor x \rfloor \, e^{-x}.$$2015-01-14
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    @dragon: it's not abuse of notation. It's clear that everything to the right having an $x$ is operated on by the integration operator. For example, what would be the context of exposing variables in $x$ as you have laid out? In fact, what rule of integration specifies that the $dx$ be at the end of the integrand?2015-01-14
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    Yeah, @dragon this is not too uncommon, and definitely not "wrong."2015-01-14
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    @dragon: It's also very common in physics.2015-01-14
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    I prefer the dx at the end as it gives some indication of 'scope'.2015-01-14
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    Could you say a few words on finding the closed form for $\sum ke^{-k}$?2018-09-11
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This reduces to a series $\displaystyle \sum_{n=0}^{\infty} \int_n^{n+1}\!\! n e^{-x}\;dx$. The integrals are easy to evaluate and so is the series.

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    But is the Riemann integral infinitely additive?2015-01-14
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    @PassingBy, some times it is. But in this case there is no need to worry about that: just use the definition of improper integral and infinite series.2015-01-14