I'm supposed to calculate the Riemann–Stieltjes integral $\int_{[0,4]} f(s)\,dZ(s)$ for $i\in(0,1)$, $f(s) = (1+i)^{\lfloor s \rfloor}$ and $Z(s) = \begin{cases} s^2, & 0 \le s \lt 1\\ 2, & 1 \le s \lt 2\\ s, & \text{otherwise} \end{cases}$
I have essentially no experience with this so here is my attempt:
\begin{align} & \int_{[0,4]} f(s)\,dZ(s) \\[8pt] = {} & \int_{[0,1)} f(s)\,dZ(s) + \int_{[1,2)} f(s)\,dZ(s) \\[4pt] & {} + \int_{[2,3)} f(s)\,dZ(s) + \int_{[3,4]} f(s)\,dZ(s) \\[8pt] = {} & \int_{[0,1)} 1 \cdot 2s \, ds + \int_{[1,2)} (i+1) \cdot 0 \, ds \\[4pt] & {} +\int_{[2,3)} (i+1)^2 \, ds + \int_{[3,4]}(i+1)^3\,ds \\[8pt] = {} & 1 +0+(i+1)^2 + (i+1)^3 \end{align}
Is that right?
EDIT:
My new attempt is as follows:
Let $X(s) := \begin{cases} s^2, & 0 \le s \lt 1\\ 1, & 1 \le s \lt 2\\ s-1, & \text{otherwise} \end{cases}$
Then $X$ is continous and we have \begin{align} & \int_{[0,4]} f(s)\,dX(s) \\[8pt] = {} & \int_{[0,1)} f(s)\,dX(s) + \int_{[1,2)} f(s)\,dX(s) \\[4pt] & {} + \int_{[2,3)} f(s)\,dX(s) + \int_{[3,4]} f(s)\,dX(s) \\[8pt] = {} & \int_{[0,1)} 1 \cdot 2s \, ds + \int_{[1,2)} (i+1) \cdot 0 \, ds \\[4pt] & {} +\int_{[2,3)} (i+1)^2 \, ds + \int_{[3,4]}(i+1)^3\,ds \\[8pt] = {} & 1 +0+(i+1)^2 + (i+1)^3 \end{align}
Now let $Y(s) := \begin{cases} 0, & 0 \le s \lt 1\\ 1, & \text{otherwise} \end{cases}$
Then we have $Z(s) = X(s) + Y(s)$ and \begin{align} & \int_{[0,4]} f(s)\,dY(s) = \int_{[0,4]} f(s)\,\delta_1(ds) = 1+i \end{align}
(I'm pretty sure up until here this should be right, the part I am really unsure about comes now:)
$\therefore$ \begin{align} & \int_{[0,4]} f(s)\,dZ(s) = \int_{[0,4]} f(s)\,d(X(s)+Y(s)) = \int_{[0,4]} f(s)\,dX(s) + \int_{[0,4]} f(s)\,dY(s) \\[8 pt] = {} & 1 +(i+1)^2 + (i+1)^3 + 1+i \end{align}
Is this now correct? Can I split up the integrator like that?