Calculate Riemann Stieltjes Integral I need help in calculating this RS integral:
$$\int_2^5[5-x] \ \ \mathrm{d}\log[x] = \log 3.$$
Here [ ] denotes greatest integer function.
This is an exercise sum of elelments of real analysis by Raisinghania, sum $1$(vi). This is the link to this question in googlebook.
Thanks in advance! :) (Precisely I need help with integration of greatest integer function.)
 A: HINT
I assume $\log x$ in your question denotes $\ln x$.
In your case where everything is nicely differentiable, you have
$$
\int_a^b f(x) dg(x) = \int_a^b f(x) g'(x) dx,
$$
so your integral ends up
$$
\int(5-x) d\ln (x) = \int \frac{5-x}{x} dx
$$
Can you finish this?
UPDATE
It seems you are interested in integrating
$$
\int \frac{[5-x]}{x}dx,
$$
where $[x]$ denotes the greatest integer function. For our example we will work with integrating $\int_a^b h(x)dx$ where $h(x) = [x]$. You can then use the same technique in your specific problem.
In case of the interval $(a,b)$, let $n = [b] - [a]$ and partition
$$
I = \int_a^b [x]dx
  = \int_a^{[a]+1} [x]dx + \int_{[a]+1}^{[a]+2} [x]dx + \ldots
                         + \int_{[a]+n}^{b} [x] dx
$$
and on each of the intervals, $[x]$ is now constant, so you get
$$
\begin{split}
I &= \int_a^{[a]+1} [a]dx + \int_{[a]+1}^{[a]+2} ([a]+1)dx + \ldots
                         + \int_{[a]+n}^{b} ([a]+n) dx \\
  &= ([a]+1-a)[a] + ([a]+1) + ([a]+2) + \ldots + (b - [b])[b]\\
  &= ([a]-a)[a] + (b -[b])[b] + n[a] + \sum_{k=1}^{n-1} k\\
  &= ([a]+n-a)[a] + (b -[b])[b] + \frac{n(n-1)}{2}.
\end{split}
$$
UPDATE 2
With the new edit of your question, you are looking to compute
$$
\int_2^5 [5-x] d\log[x],
$$
with the greatest-integer function in the log under differentiation. Since $[x]$ and hence $\log [x]$ is constant almost everywhere, you cannot use the shorthand for derivatives I quoted above. The idea is that $\Delta g(x) = 0$ everywhere except 3 points at $x \in \{3,4,5\}$. In each of these places, you have $f(x) = [5-x] \in \{2, 1, 0\}$ and $\Delta g(x) = \Delta \ln [x] \in \{ \ln 3 - \ln 2, \ln 4 - \ln 3, \ln 5 - \ln 4\}$. Hence the integral becomes
$$
\begin{split}
\int_2^5 [5-x] d\log[x]
 &= \sum_k f_k (\Delta g)_k \\
 &= 2 (\ln 3 - \ln 2) + 1 (\ln 4 - \ln 3) + 0 (\ln 5 - \ln 4) \\
 &= 2 \ln 3 - 2 \ln 2 + \ln 4 - \ln 3 \\
 &= \ln 3,
\end{split}
$$
since $2\ln 2 = \ln 2^2 = \ln 4$.
