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Calculate: $\int_0^\infty [x]e^{-x} \, dx$ where $[x]:=\max \{k\in\mathbb{Z}:k\leq x\}$

Solution: $$\int_0^\infty[x]e^{-x} \, dx = \sum_{k=0}^\infty \int_k^{k+1}[x]e^{-x} \, dx = \sum_{k=0}^\infty k\int_k^{k+1}e^{-x} \, dx = \sum_{k=0}^\infty k(-e^{-1}+1)e^{-k}$$

It is right? How can I solve the series? Thank you.

Michael Hardy
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Andrés Villa
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    $$ S = \sum_{k=0}^{\infty} k e^{-k} = \sum_{k=1}^{\infty} k e^{-k} = \sum_{k=0}^{\infty} \frac {k+1} e e^{-k} = \frac 1e \sum_{k=0}^{\infty} k e^{-k} + \frac 1 e \sum_{k=0}^{\infty} e^{-k} = \frac Se + \frac 1 e \sum_{k=0}^{\infty} e^{-k}$$ – Dmitry Aug 07 '20 at 03:50
  • In recent decades it has become standard to denote $\max { k \in \mathbb{Z} : k\leq x}$ by $\lfloor x\rfloor$ rather than by $[x]. \qquad$ – Michael Hardy Aug 07 '20 at 04:11
  • Possible duplicate of https://math.stackexchange.com/q/3226580/478779 – Paras Khosla Aug 07 '20 at 05:01

4 Answers4

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You're almost done. Note that $$\begin{aligned} \int_0^{\infty} \lfloor x \rfloor e^{-x} \; \mathrm{d}x &= \sum_{k=0}^{\infty} \int_k^{k+1} ke^{-x} \; \mathrm{d}x\\ &= \sum_{k=0}^{\infty} \left [-ke^{-x} \right ]_k^{k+1}\\ &= \sum_{k=0}^{\infty} k\left (e^{-k} - e^{-k-1}\right )\\ &= \sum_{k=0}^{\infty} ke^{-k} - \sum_{k=0}^{\infty} ke^{-k-1}\\&= \sum_{k=1}^{\infty} ke^{-k} - \sum_{k=1}^{\infty} (k-1)e^{-k}\\ &= \sum_{k=1}^{\infty} e^{-k}\\ &= \frac{e^{-1}}{1 - e^{-1}}\\ &= \frac{1}{e-1}. \end{aligned}$$ Hence, $$\int_0^{\infty} \lfloor x \rfloor e^{-x} \; \mathrm{d}x = \frac{1}{e-1}.$$

Sharky Kesa
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Hint: For any $|z|<1$

$\frac{1}{(1-z)^2} = \sum_\limits{n\geq1}nz^{n-1}$ and so,

$\frac{z}{(1-z)^2}=\sum_\limits{n\geq1}nz^n$

Mittens
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To evaluate the final sum, rewrite it as a telescoping sum as follows: $$\sum_{k=0}^{\infty} ke^{-k}(1-e^{-1}) = \sum_{k=0}^{\infty} ke^{-k} -ke^{-(k+1)} $$

Writing out the first couple of terms: $$ 0-0 + \frac{1}{e} - \frac{1}{e^2}+\frac{2}{e^2}-\frac{2}{e^3}+\frac{3}{e^3}-\frac{3}{e^4} = \frac{1}{e}+\frac{1}{e^2}+\frac{1}{e^3}+... $$

This is an infinite geometric series and $\frac{1}{e}$ is within the radius of convergence, that is $\big|\frac{1}{e}\big|\le 1$. It is, however, missing the first term, $1$.

$$\sum_{k=0}^{\infty}r^k=\frac{1}{1-r} $$

Setting $r=\frac{1}{e}$ and subtracting $1$, the sum is now $$\frac{1}{1-\frac{1}{e}}-1=\frac{e}{e-1}-\frac{e-1}{e-1}=\frac{1}{e-1} $$

C Squared
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Here's as general a generalization as I can come up with.

What is $K(a) =\int_0^\infty \lfloor f(x)\rfloor h(x)dx $ where $f(0) = 0, f'(x) > 0, g(f(x) = f(g(x) = x, h(0) > 0, h'(x) < 0, h(x) \to 0, H'(x) = h(x) $ as $x \to \infty$?

$\begin{array}\\ K(a) &=\int_0^\infty \lfloor f(x)\rfloor h(x)dx\\ &=\sum_{m=0}^{\infty}\int_{f(x) \ge m}^{f(x) \le m+1} mh(x)dx\\ &=\sum_{m=0}^{\infty}m\int_{g(n)}^{g(m+1)} h(x)dx\\ &=\sum_{m=0}^{\infty}m(H(x)|_{g(m)}^{g(m+1)})\\ &=\sum_{m=0}^{\infty}m(H(g(m))-H(g(m+1))\\ &=\sum_{m=0}^{\infty}mH(g(m))-\sum_{m=0}^{\infty}mH(g(m+1))\\ &=\sum_{m=0}^{\infty}mH(g(m))-\sum_{m=1}^{\infty}(m-1)H(g(m))\\ &=\sum_{m=0}^{\infty}mH(g(m))-\sum_{m=1}^{\infty}mH(g(m))+\sum_{m=1}^{\infty}H(g(m))\\ &=\sum_{m=1}^{\infty}H(g(m))\\ \end{array} $

If $h(x) = e^{-x}, f(x) = x $ then $g(x) = x, H(x) = e^{-x} $ so the result is $\sum_{m=1}^{\infty}H(g(m)) =\sum_{m=1}^{\infty}e^{-m} =\dfrac{1/e}{1-1/e} =\dfrac1{e-1} $.

marty cohen
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