Using Riemann–Stieltjes to compute an integral

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, 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!

• You could use the fact that if $$I(x) = \begin{cases}0 & (x<0)\\ 1 & (x \geq 0)\end{cases},$$and if $f$ is bounded in $[a,b]$ and continuous in $s \in (a,b)$, then $$\int_a^b fdI(x-s)=f(s).$$ This can be easily demonstrated using the definition (se, e.g. "baby" Rudin, Theorem 6.15, page 130).
– dfnu
Dec 11 '20 at 19:00
• Also, you should consider, I think, more carfully the last equality in your chain. Suppose all $x_i$'s are less than $1$. Then the quantity in parenthesis after the first equality is (correctly) $0$, independently of $i$, whereas the last quantity in the chain is strictly greater than $0$.
– dfnu
Dec 11 '20 at 19:18
• Ok! Thanks! I also think about using the integration by parts to tackle it and write it in the below. Could you please give me some suggestions? $$\int_0^b x d (F(x))=b \sum_{k=0}^{[b]}\frac{2^k}{k!}-\int_0^b\sum_{k=0}^{[x]}\frac{2^k}{k!}dx,$$ where $F(x)=\sum_{k=0}^{[x]}\frac{2^k}{k!}$. I am not sure how to evaluate the last integral. Thanks again! Dec 11 '20 at 19:43
• Did you sketch the graph of $F(x)$? It may give you some other hints.
– dfnu
Dec 11 '20 at 20:18
• I would also check the hypotheses under which integration by parts is allowed.
– dfnu
Dec 11 '20 at 20:38