# The Unapologetic Mathematician

## The Riemann-Stieltjes Integral II

What with cars not working right, I’m not along with the undergraduate math club at the MAA meeting in Lake Charles. That honor goes to Steve Sinnott alone. And so I’m still back in New Orleans, and I may as well write another post today.

Let’s follow on yesterday’s discussion of the Riemann-Stieltjes integral by looking at a restricted sort of integrator. We’ll assume here that $\alpha(t)$ is continuously differentiable — that is, that $\alpha'(t)$ exists and is continuous in the region we care about. We’ll also have a function $f$ on hand to integrate.

Now let’s take a tagged partition $x=((x_0,...,x_n),(t_1,...,t_n))$ and set up our Riemann-Stieltjes sum for $f$

$\displaystyle f_{\alpha,x}=\sum\limits_{i=1}^nf(t_i)(\alpha(x_i)-\alpha(x_{i-1}))$

We can also use the partition to set up a Riemann sum for $f\alpha'$

$\displaystyle(f\alpha')_x=\sum\limits_{i=1}^nf(t_i)\alpha'(t_i)(x_i-x_{i-1})$

Now the Differential Mean Value Theorem tells us that in each subinterval $\left[x_{i-1},x_i\right]$ there is some $s_i$ so that $(\alpha(x_i)-\alpha(x_{i-1})=\alpha'(s_i)(x_i-x_{i-1})$. We stick this into the Riemann-Stieltjes sum and subtract from the Riemann sum

$\displaystyle(f\alpha')_x-f_{\alpha,x}=\sum\limits_{i=1}^nf(t_i)(\alpha'(t_i)-\alpha'(s_i))(x_i-x_{i-1})$

And continuity of $\alpha'$ tells us that as we pick the partitions finer and finer, we’ll squeeze $s_i$ and $t_i$ together, and so $\alpha'(t_i)-\alpha'(s_i)$ goes to zero. So, when the integrator $\alpha$ is continuously differentiable, the Riemann-Stieltjes integral reduces to a Riemann integral

$\displaystyle\int\limits_{[a,b]}f(x)d\alpha(x)=\int\limits_{[a,b]}f(x)\alpha'(x)dx$

Now let’s go back to the fence story from yesterday. There’s really some function $g(p)$ that tells us the height of the fence at position $p$. We got our integrator from the function $p=\alpha(t)$, which takes a time $t$ and gives the position $p$ at that time. Thus our function giving height at time $t$ is the composition $f(t)=g(\alpha(t))$. We also note that as time goes from $a$ to $b$, we walk from $\alpha(a)$ to $\alpha(b)$. If we put all this into the above equation we find

$\displaystyle\int\limits_{[\alpha(a),\alpha(b)]}g(p)dp=\int\limits_{[a,b]}g(\alpha(t))\alpha'(t)dt$

which is just our change of variables formula.

So here’s what this formula means: we have a real-valued function on some domain — here it’s the height of the fence as a function of position along the ground. We take a region in the space of these variables — the section of the fence we’re walking past — and we “parametrize” it by describing it with a (sufficiently smooth) function taking a single real variable — the position function $p=\alpha(t)$. We “pull back” our function by composing it with this parametrization, giving us a real-valued function of a real variable, which we can integrate.

The deep thing here is that we have two different parametrizations. We can describe position as $p$ meters from a fixed starting point, or we can describe it as the point we’re passing at $t$ seconds from a certain starting time. In fact, as we change our function $\alpha$ we get all sorts of different parametrizations of the same stretch of ground. And we should measure the same area for the same fence no matter which parametrization we choose.

Given a parametrization $\alpha$, the integral in our recipe will be the Riemann-Stieltjes integral

$\displaystyle\int\limits_{[a,b]}g(\alpha(t))d\alpha(t)$

which we can reduce to the Riemann integral

$\displaystyle\int\limits_{[a,b]}g(\alpha(t))\alpha'(t)dt$

and it will be the same answer as if we used the “natural” parametrization

$\displaystyle\int\limits_{[\alpha(a),\alpha(b)]}g(p)dp$

And that’s why the change of variables works.

February 29, 2008 Posted by | Analysis, Calculus | 15 Comments