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I think I understand these concepts ok:

But there seems to be a layer that I miss: They all seem to be connected, alas I don't know how exactly. In a way you should be able to transform one into the other?!?

Could anyone please enlighten me? Thank you!

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See… – lhf Apr 5 '11 at 13:00
@lhf: Thank you, I found that already - I think it gives only part of the answer and not the big picture. – vonjd Apr 5 '11 at 13:02
up vote 14 down vote accepted

The Chain Rule and $u$-substitution are "inverses of each other." In the Chain Rule, you obtain the derivative of a composition: $$\Bigl(f(g(x))\Bigr)' = f'(g(x))g'(x).$$

In $u$-substitution, you "recognize" a product of two functions $h(x)g(x)$ as an instance of $f'(g(x))g'(x)$ by doing a substitution. So $$\frac{2x}{x^2+1} = \frac{1}{x^2+1}(2x)$$ is "recognized" as $f'(g(x))g'(x)$ by taking $g(x) = x^2+1$, which makes $f'(u) = \frac{1}{u}$.

That is, the connection between the Chain Rule and $u$-substitution is completely and absolutely direct; one "undoes" what the other one does. The $u$-substitution simplifies the result of the Chain Rule to make integration "more obvious", calling $g(x)=u$ and $g'(x)\,dx = du$, so that instead of $f'(g(x))g'(x)\,dx$ you have $f'(u)\,du$.

The connection with Riemann-Stieltjes integral is rather more tenuous. Remember that the Riemann-Stieltjes integral of $f(x)$ with respect to $g(x)$, $$\int_a^b f(x)dg(x)$$ is defined to be the limit over partitions $P$ of $[a,b]$ as the mesh size goes to zero of the Riemann-Stieltjes sum $$S(P,f,g) = \sum_{i=0}^{n-1} f(x_i^*)(g(x_{i+1})-g(x_i)),$$ where $x_i^*$ is an arbitrary point in the partition interval $[x_i,x_{i+1}]$.

When $g$ is differentiable, the Mean Value Theorem tells us that there exists $x'_i$ in each $(x_i,x_{i+1})$ such that $g'(x'_i)(x_{i+1}-x_i) = g(x_{i+1})-g(x_i)$, so that we can replace the Riemann-Stieltjes sum (by switching, if necessary, the point $x_i^*$ of evaluation of $f$ to the same point $x'_i$) with: $$\sum_{i=0}^{n-1} f(x'_i)g'(x'_i)(x_{i+1}-x_i)$$ which is a Riemann sum for the function $h(x)=f(x)g'(x)$. Therefore, taking the limit over the partitions $P$ of $[a,b]$ as the mesh size goes to zero of this Riemann sum gives the integral of $h(x)$, and so you get the equality

$$\begin{align*} \int_a^b f(x)\,dg(x) &= \lim_{||P||\to 0} \sum_{i=0}^{n-1}f(x_i^*)(g(x_{i+1})-g(x_i))\\ & = \lim_{||P||\to 0}\sum_{i=0}^{n-1}f(x'_i)g'(x'_i)(x_{i+1}-x_i) = \int_a^b f(x)g'(x)\,dx.\end{align*}$$

The integral on the right is similar, but not equal, to what you get with the Chain Rule; in the Chain Rule you would have $f(g(x))g'(x)$, rather than $f(x)g'(x)$. Instead, it's more like "half of a product rule" ( $(fg)' = fg' + f'g$, so here you have that first summand but not the second). So while the Riemann-Stieltjes integral looks somewhat like the Chain Rule (when $g(x)$ is differentiable), it's not quite the same.

Of course, it's possible that when you do some Riemann-Stieltjes integrals, particularly in a classroom setting, the choices of $f$ and $g$ will be precisely such that the simplified integral $\int f(x)g'(x)\,dx$ just happens to be doable with a $u$-substitution; e.g., it "just happens" that $f(x)g'(x) = \mathcal{F}'(h(x))h'(x)$ for a function $h(x)$ that has the same derivative as $g(x)$, and a function $\mathcal{F}$ that has derivative similar to $f$. For instance, if $f(x) = \frac{1}{1+x^2}$, $g(x) = x^2$, $h(x)=1+x^2$ (note that $g'(x)=h'(x)$) and $\mathcal{F}'(x) = \frac{1}{x}$, we get that $$f(x)g'(x) = \frac{1}{1+x^2}(x^2)' = \frac{1}{1+x^2}(1+x^2)' = \mathcal{F}'(h(x))h'(x) = \Bigl(\mathcal{F}(h(x))\Bigr)'$$ where $\mathcal{F}(x) = \ln|x|+C$ (since $\mathcal{F}'(x) = \frac{1}{x}$).

So doing a lot of Riemann-Stieltjes integration problems where the choices are selected to make this happen may give the illusion that Riemann-Stieltjes integration is also connected to the Chain Rule/$u$-substitution, but that is likely to be an artifact of the problems you are asked to solve, rather than a direct connection between them.

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+1: for a detailed and interesting answer. – Fabian Apr 5 '11 at 16:20
Thank you for this very enlightening answer! – vonjd Apr 5 '11 at 18:31

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