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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.

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    I think you are right, nice insight! For example, let $f(x)=x^{3/2}$, and we are interested in $\int f(x)\,dx$. (It is a definite integral, but I don't want to worry about the limits). Let $x=e^u$. The change of variable formula says our integral is $\int f(e^u)d(e^u)$, which can be viewed as a Stieltjes integral.2012-02-02

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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 $\dfrac {u_{n+1}}{ u_{n}}=\rm const.$

$\Delta a_n=a_{n+1}-a_n=\log u_{n+1} - \log u_{n}=\log \dfrac {u_{n+1}}{ u_{n}}=\log c =\rm const.$