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

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:

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

• By the definition of Riemann integration, one is always integrating on some interval $[a,b]$. And it can be proved that if $f$ is Riemann integrable, then $f$ is bounded. "Improper" integrals merely generalize Riemann integration. – florence Jul 2 '16 at 5:35
• @florence so how does one explicitly write $\int_0^\infty f(x) dx$ as a Riemann sum? – drewdles Jul 2 '16 at 5:40
• What one does is defining $\int_0^{\infty} f dx$ as the limit $\lim_{\omega \to \infty} \int_0^{\omega} f dx$. – user305860 Jul 2 '16 at 5:43
• @ErikJoensson I slightly edited the question in response to your comment. Hopefully it is clearer as to what I'm asking ... – drewdles Jul 2 '16 at 5:49

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.

• So So $\lim_{k \to \infty}\lim_{n \to \infty} \sum_{r=1}^n f( \frac{kr}{n}) \frac{k}{n} = \int_{0}^\infty f(x) dx$ makes sense? – drewdles Jul 2 '16 at 5:58
• yes, this is should work. (the inner limit is the rieman integral over $[0,k]$ and the outer limit turns it into an improper integral) EDIT: as far as i know, riemann-integrals are defined by refinements. note that the series of intervals with length $\frac{k}{n}$ gives not a refinement (but of course contains a whole lot and i think the limit of your expression is still the riemann integral) – Max Jul 2 '16 at 6:10
• Well, things can get a bit complicated. For instance, if you have a singularity at zero, you might need to take $$\lim_{\varepsilon\to 0^+} \int_\varepsilon^{1/\varepsilon} f(x).$$ But for many functions $f$ the thing you wrote down is the "right" choice for an improper integral. (I'm not aware of any precise definition for what the improper integral notation means; usually it's obvious from context.) – Daniel McLaury Jul 2 '16 at 6:13

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.

• So $\lim_{k \to \infty}\lim_{n \to \infty} \sum_{r=1}^n f( \frac{kr}{n}) \frac{k}{n} = \int_{0}^\infty f(x) dx$ is senseless? – drewdles Jul 2 '16 at 5:54
• It would seem so, what is that k doing there? It doesn't make sense to me. – DpS Jul 2 '16 at 5:58
• Umm ... Well just as we have $\int_{a}^b f(x) dx = \lim_{n \to \infty} \sum_{r=1}^n f(\frac{(b-a)r}{n}) \frac{b-a}{n}$ and then taking limit both sides and setting $a$ to 0: $\int_{0}^\infty f(x) dx = \lim_{b \to \infty} \lim_{n \to \infty} \sum_{r=1}^n f(\frac{(b)r}{n}) \frac{b}{n}$ – drewdles Jul 2 '16 at 6:04
• This looks fine for a finite b, but wouldn't b/n be a problem if both b and n go to infinity? – DpS Jul 2 '16 at 6:08