I am calculating the following Riemann–Stieltjes integral
$$\int_0^bx \ d\left(\sum_{k=0}^{[x]}\frac{2^{k}}{k!}\right)$$
where $x>0$ and $[\,\cdot\,]$ is the floor function.
My idea is using the definition to write $$\int_0^bx \, d\left( \sum_{k=0}^{[x]} \frac{2^{k}}{k!} \right) = \lim_{n\rightarrow\infty} \sum_{i=1}^n x_i\ \left(\sum_{k=0}^{[x_i]}\frac{2^k}{k!}-\sum_{k=0}^{[x_{i-1}]} \frac{2^k}{k!}\right) = \lim_{n\rightarrow\infty}\sum_{i=1}^n x_i\frac{2^{[x_i]}}{[x_i]!},$$ here I use a partition $0= x_1<\cdots<x_n=b$, but I have no idea to tackle the last part I got.
My second trying is using the integration by parts and obtain $$\int_0^b x \ d\left(\sum_{k=0}^{[x]}\frac{2^{k}}{k!}\right)=b \sum_{k=0}^{[b]}\frac{2^k}{k!}-\int_0^b\sum_{k=0}^{[x]}\frac{2^k}{k!}dx.$$ I am not sure how to evaluate the last integral.
Any suggestion would be welcome! Thanks!