How do you resolve $\int x (dx)^2$? The notes I am seeing says the term goes to zero, but I don't understand how. Please help.
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$\begingroup$ Welcome to Math SE. Could you be a bit more explicit? $\endgroup$– mflNov 15, 2014 at 21:05
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2 Answers
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Let us interpret $\int_a^b x(dx)^2$ as a limit of Riemann sum. We have $$\int_a^b x(dx)^2 = \lim_{n \to \infty}\sum_{k=1}^{n}\left(a+k\cdot\dfrac{b-a}n\right)\left(\dfrac{b-a}n\right)^2 = \lim_{n \to \infty} \left(an+\dfrac{n(n+1)(b-a)}{2n}\right)\left(\dfrac{b-a}n\right)^2$$ Hence, we obtain $$\int_a^b x(dx)^2 = \lim_{n \to \infty} \left(\dfrac{a(b-a)^2}n+\dfrac{(n+1)(b-a)^3}{2n^2}\right) = 0$$ This is true with any higher order power of $(dx)$, i.e., if $\displaystyle \int_a^b f(x) dx \in \mathbb{R}$, then $\displaystyle \int_a^b f(x) (dx)^p$, where $p>1$ is zero.
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$\begingroup$ Very succinct, this helped me understand. Did not even occur to me to represent as Riemann sum. Thank you. $\endgroup$ Nov 15, 2014 at 21:15
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$\begingroup$ I think this argument is invalid. The Riemann integral does not include a general substitution of $dx$ into $\Delta x/n$ for whatever term behind a summation $\int$ symbol. I could claim (fishy as well) that $\int x (dx)^2 = ((1/2) x^2 + C) dx$. $\endgroup$– mvwNov 15, 2014 at 21:33
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$\begingroup$ I think you would be right with that claim, but when you resolve the integral (plug in a and b), the dx term will equal zero, thus the whole equation will equal zero. This is what I think the Riemann sum approach was trying to illustrate. $\endgroup$ Nov 17, 2014 at 23:45
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You need to assign a meaning to your intended calculation.
What does $x (dx)^2$ mean in the first place? Is it the same as $x\, dx \,dx$?
What kind of integration is exactly meant if you add an summation $\int$ symbol to your term? What input mathematical objects what output objects what summation procedure?
There is no useful integration for this kind of term defined that I am aware of (maybe some non-standard analysis folk has one).
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$\begingroup$ Yes, it does mean x dx dx. The reason I am asking is that I was looking at a proof that used $\int (x^2 + x dx + dx dx)dx = x^3/3$. I found three different sources with the same proof that gave the same answer. I just didn't understand how they got the answer. $\endgroup$ Nov 15, 2014 at 21:58
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$\begingroup$ Looks like 17th century style infinitesimal calculus. So what kind of integral is that supposed to be? Riemann? $\endgroup$– mvwNov 15, 2014 at 22:07