Q:

Interpreting an integral under the Riemann Stieltjes form.

I've long ago solved an exercise that was as follows:
This problem provides a new way of finding $ \displaystyle \int\limits_a^b {{x^p}dx} $ for $ 0 < a < b $. It consists of using partitions $t_i$ for which the quotient $\displaystyle \frac{t_i}{t_{i-1}}$ is constant, instead of being $ t_i - t_{i-1} $ constant.
The solutions produced $ t_i = a c^{\frac{i}{n}} $ where $ c = \frac{b}{a} $ and upper and lower sums being:
$$ U \left( f, P \right) = \left( {{b^{p + 1}} - {a^{p + 1}}} \right)\frac{{{c^{\frac{p}{n}}}}}{{1 + {c^{\frac{1}{n}}} + {c^{\frac{2}{n}}} +  \cdots  + {c^{\frac{p}{n}}}}} $$
$$L\left( f, P \right) = \left( {{b^{p + 1}} - {a^{p + 1}}} \right)\frac{1}{{1 + {c^{\frac{1}{n}}} + {c^{\frac{2}{n}}} +  \cdots  + {c^{\frac{p}{n}}}}}$$
Thus you have $$U \left( f, P \right) -  L \left( f, P \right) = \left( {{b^{p + 1}} - {a^{p + 1}}} \right)\left[ {\frac{{{c^{\frac{p}{n}}} - 1}}{{1 + {c^{\frac{1}{n}}} + {c^{\frac{2}{n}}} +  \cdots  + {c^{\frac{p}{n}}}}}} \right]$$
which implies that for some $\epsilon > 0$ and $N$ sufficiently large.
$$ U \left( f, P \right) -  L \left( f, P \right) < \epsilon $$
Finally you get the expected result 
$$\int\limits_a^b {{x^p}dx}  = \frac{{{b^{p + 1}} - {a^{p + 1}}}}{{p + 1}} $$
I' m wondering if this change in the differential $ t_i - t_{i-1} = \Delta{t_i}$ versus $\displaystyle \frac{t_i}{t_{i-1}}= \Delta{t_i}$ can be interpreted as changing the integrator $d\alpha$. I've started reading about the Riemann Stieltjes integration so it rang a bell, but don't expect me to know a lot about it, just the basics.

A:

Just for the sake of closing, the integrator would we
$$\log x$$ so one has
$$\int_a^b x^{p} dx=\int_a^b x^{p+1} d({\log x})$$
as Riemann integral.
Thus with $a_n=\log u_{n}$ and $\dfrac {u_{n+1}}{ u_{n}}=\rm const.$
$$\Delta a_n=a_{n+1}-a_n=\log u_{n+1} - \log u_{n}=\log \dfrac {u_{n+1}}{ u_{n}}=\log c =\rm const.$$

