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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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    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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    @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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    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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    @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