Calculate: $\int_0^\infty [x]e^{-x} \, dx$ where $[x]:=\max \{k\in\mathbb{Z}:k\leq x\}$ 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.
 A: 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,
$$\int_0^{\infty} \lfloor x \rfloor e^{-x} \; \mathrm{d}x = \frac{1}{e-1}.$$
A: Hint:
For any $|z|<1$
$\frac{1}{(1-z)^2} = \sum_\limits{n\geq1}nz^{n-1}$ and so,
$\frac{z}{(1-z)^2}=\sum_\limits{n\geq1}nz^n$
A: 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 $$\frac{1}{1-\frac{1}{e}}-1=\frac{e}{e-1}-\frac{e-1}{e-1}=\frac{1}{e-1} $$
A: 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}
$.
