How to calculate this Riemann-Stieltjes integral?

Question

How to calculate this Riemann-Stieltjes integral?

\begin{equation} \int_{1}^{3}e^{x}d\left\lfloor x\right\rfloor \end{equation}

If $x\in\left[0,\,3\right]$, then $\left\lfloor x\right\rfloor =I\left(x-1\right)+I\left(x-2\right)+I\left(x-3\right)$, where $I\left(x-a_{i}\right)=0$ if $x<a_{i}$, $I\left(x-a_{i}\right)=1$ if $x\geq a_{i}$.

Then it is easy, we have $\int_{0}^{3}e^{x}d\left\lfloor x\right\rfloor =\sum_{n=1}^{3}e^{a_{i}}$.

But here $x\in\left[1,\,3\right]$, so how to deal with it ? Thank you very much.

Answer 1

Let $$f(x) = e^x, \;g(x) = \lfloor x \rfloor, \\ I = \int_1^3 f(x) dg(x).$$ If you have a formula to compute $F(x) = \int_0^x f(t) dg(t)$, then $I = F(3) - F(1)$.

Alternatively, $$I = f(2) \lim_{\epsilon \downarrow 0} (g(2 + \epsilon) - g(2 - \epsilon)) + f(3) \lim_{\epsilon \downarrow 0} (g(3) - g(3 - \epsilon)).$$

As another option, $$I = f(3) g(3) - f(1) g(1) - \int_1^3 f'(x) g(x) dx.$$

Answer 2

I'm about five years late to this one, but since it's still unanswered I figured I might as well post an answer to it anyways.

So in general, it's the case that $$\int_1^nf(x)\ d \left \lfloor x \right \rfloor = \sum _{k=2} ^n f(k).$$ Applying this identity here gives that $$\int _1 ^3 e^x \ d \left \lfloor x \right \rfloor = \sum_{k = 2} ^3e^k = e^2 + e^3.$$ To prove the above identity, I would suggest (as a small hint) splitting the function $\left \lfloor x \right \rfloor$ into three step functions.