Why can Riemann integrals can be computed only for proper integrals?

Question

I was reading http://mathworld.wolfram.com/RiemannIntegral.html where at the end it says "Riemann integrals can be computed only for proper integrals" where the definition of a proper integral is: "An integral which has neither limit infinite." Why can't one take:

$$ \lim_{b \to \infty} \int_{a}^b f(x) dx $$

and evaluate these kinds of integrals as well?

But at the same time I seem to be confused how to write the integral explicitly as a limit of a sum/Riemann sum (with the order of the limits). For example how does one write $\int_0^\infty f(x) dx$? Is it:

$$ \lim_{k \to \infty}\lim_{n \to \infty} \sum_{r=1}^n f( \frac{kr}{n}) \frac{k}{n} $$

Answer 1

Well, see there's the problem. You can't partition an unbounded interval into n subintervals of equal length. Hence you can't define the Reimann sum over an unbounded interval, therefore Reimann integrals can be computed only for proper integrals.

Answer 2

Why can't one take: $ \lim\limits_{b \to \infty} \int_{a}^b f(x) dx $ and evaluate these kinds of integrals as well?

You can (and do); it's just that that thing isn't itself a Riemann integral.

In order to take a Riemann integral, you need to divide the domain up into a finite number of finite-length pieces, which is impossible to do if the domain has infinite length.

Since the limit of Riemann integrals doesn't satisfy the definition of a Riemann integral, it isn't a Riemann integral. In particular, this means that if you prove a theorem about Riemann integrals, it may or may not apply to these limit-of-Riemann-integral gadgets.

One example of this phenomenon is Lebesgue's result that the Riemann-integrable functions are precisely the functions which are continuous away from a set of measure zero, and that the Lebesgue integral of such a function is equal to the Riemann integral. Since this theorem uses the definition of the Riemann integral, it (a priori) only applies in that case, and in fact it turns out that there are functions with improper (Riemann) integrals which are not Lebesgue integrable.

To be fair, if it had turned out to be the case that pretty much every theorem which applied to Riemann integrals also applied to improper integrals, we probably would have altered the definition of the Riemann integral to encompass these integrals. Since they have somewhat different properties, though, we keep them separate.