The Unapologetic Mathematician

Mathematics for the interested outsider

The Riemann-Stieltjes Integral IV

Let’s do one more easy application of the Riemann-Stieltjes integral. We know from last Friday that when our integrator is continuously differentiable, we can reduce to a Riemann integral:

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

So where else have we seen derivatives as factors in integrands? Right! integration by parts! Here our formula says that

\displaystyle\int\limits_a^bf(x)g'(x)dx+\int\limits_a^bf'(x)g(x)dx=f(b)g(b)-f(a)g(a)

We can rewrite this using Riemann-Stieltjes integrals as

\displaystyle\int\limits_{\left[a,b\right]}f(x)dg(x)+\int\limits_{\left[a,b\right]}g(x)df(x)=f(b)g(b)-f(a)g(a)

So if f and g are both continuously differentiable, this formula gives back our rule for integration by parts. But we can prove this without making those assumptions. In fact, we just need to assume that one of the two integrals exists, and the existence of the other one (and the formula) will follow.

Let’s assume that \int_{\left[a,b\right]}f,dg exists. That is, for every \epsilon>0 there is some tagged partition x_\epsilon so that for every finer partition x we have

\displaystyle\left|\int\limits_{\left[a,b\right]}f(x)dg(x)-f_{g,x}\right|<\epsilon

Now let’s take any partition x finer than x_\epsilon and use it to set up the Rieman-Stieltjes sum

\displaystyle g_{f,x}=\sum\limits_{i=1}^ng(t_i)f(x_i)-\sum\limits_{i=1}^ng(t_i)f(x_{i-1})

We can also use this partition to rewrite A=f(b)g(b)-f(a)g(a) as

\displaystyle A=\sum\limits_{i=1}^nf(x_i)g(x_i)-\sum\limits_{i=1}^nf(x_{i-1})g(x_{i-1})

So subtracting the one from the other we find

\displaystyle A-g_{f,x}=\sum\limits_{i=1}^nf(x_i)(g(x_i)-g(t_i))+\sum\limits_{i=1}^nf(x_{i-1})(g(t_i)-g(x_{i-1}))

But this is a Riemann-Stieltjes sum f_{g,\bar{x}} for the partition \bar{x} we get by throwing together all the x_i and t_i as partition points, and using x_i as tags. This is a finer partition than x_\epsilon, and so we see that

\displaystyle\left|\left(A-\int\limits_{\left[a,b\right]}f(x)dg(x)\right)-g_{f,x}\right|=\left|\int\limits_{\left[a,b\right]}f(x)dg(x)-\left(A-g_{f,x}\right)\right|<\epsilon

whenever x is a partition finer than x_\epsilon. This shows that the Riemann-Stieltjes integral of g with respect to f exists, and has the value we want.

March 4, 2008 Posted by | Analysis, Calculus | 1 Comment

   

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