Riemann-Stieltjes integration with floor function Evaluate $$\int_{\frac{2}{3}}^8 f(x)d\alpha(x)$$ where $\alpha$ is continuous and $f$ is the floor function, that is $f(x)$ is the greatest integer less than or equal $x$.
 A: Hint:
Write $\int_{\frac{2}{3}}^8 \lfloor x \rfloor d \alpha(x) = \int_{\frac{2}{3}}^1 \lfloor x \rfloor d \alpha(x)+ \sum_{k=1}^7 \int_{k}^{k+1} \lfloor x \rfloor d \alpha(x) $.
What value does $\lfloor x \rfloor$ take inside these integrals?
A: $$I = \int_{2/3}^8 \lfloor x\rfloor d \alpha(x) = \int_{2/3}^1 \lfloor x\rfloor d \alpha(x) + \sum_{k=1}^7 \int_k^{k+1} \lfloor x\rfloor d \alpha(x) = 0 + \sum_{k=1}^7 \int_k^{k+1} \lfloor x\rfloor d \alpha(x)$$
Now $\lfloor x \rfloor = k$ for $x \in [k,k+1)$. Hence, we get that
\begin{align}
I & = \sum_{k=1}^7 k (\alpha(k+1) - \alpha(k))\\
& = \sum_{k=2}^7 \alpha(k) (-k + k-1) - \alpha(1) + 7 \alpha(8)\\
& = 7 \alpha(8) - \sum_{k=1}^7 \alpha(k)
\end{align}

EDIT
To make the last step clear, let us explicitly write it out and see.
\begin{align}
I & = \sum_{k=1}^7 k (\alpha(k+1) - \alpha(k))\\
& = 1 \cdot(\alpha(2) - \alpha(1)) + 2 \cdot(\alpha(3) - \alpha(2)) + 3 \cdot(\alpha(3) - \alpha(2)) + 4 \cdot(\alpha(4) - \alpha(3))\\
& + 5 \cdot(\alpha(5) - \alpha(4)) + 6 \cdot(\alpha(6) - \alpha(5)) + 7 \cdot(\alpha(8) - \alpha(7))\\
& = -\alpha(1) + (1-2) \cdot \alpha(2) + (2-3) \cdot \alpha(3) + (3-4) \cdot \alpha(4) + (4-5) \cdot \alpha(5) + (5-6) \cdot \alpha(6)\\
& + (6-7) \cdot \alpha(7) + 8 \cdot \alpha(8)\\
& = -\alpha(1) - \alpha(2) - \alpha(3) - \alpha(4) - \alpha(5) - \alpha(6) - \alpha(7) + 7 \alpha(8)\\
& = 7 \alpha(8) - \sum_{k=1}^7 \alpha(k)
\end{align}
