Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You might be interested in knowing that this method of "quadrature" goes back to Fermat, before either Newton or Leibniz was born. Fermat used it to find closed forms for the area under curves $y=x^r$, where $r$ is rational. – André Nicolas Feb 1 '12 at 20:57
Ok - good to know that. Any insights on Riemann Stieltjes? – Pedro Tamaroff Feb 2 '12 at 0:45
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. – André Nicolas Feb 2 '12 at 1:51
up vote 1 down vote accepted

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.