# Functions continuous on half open intervals are Riemann Integrable?

In my textbook there's this theorem:

Let $f$ be a bounded function on $[a,b]$ that's continuous on $(a,b]$. Then the Riemann integral of $f$ exists on $[a,b]$.

But then the next page says:

Let $f$ be a continuous function on $(a,b]$. If the limit of $\int_{a+d}^bf(x)dx$ as $d$ goes to $0$ exists then we define that limit to be the improper Riemann integral of $f$. Similar definition holds if $f$ is continuous but unbounded on $[a,b)$.

By the first theorem, functions such as $\frac{sin(x)}{x}$ is Riemann integrable, but by the second theorem, it is not Riemann integrable but exists as improper integral. How do I reconcile this?

• The second theorem is valid only when the first is not – N74 May 5 '16 at 23:30
• It will depend on the exact definition of improper integral. I think of the "improperly Riemann integrable" functions on $(a,b]$ as a superset of the Riemann integrable functions. So your $\frac{\sin x}{x}$ is both Riemann integrable and improperly Riemann integrable. – André Nicolas May 5 '16 at 23:32
• the first theorem applies to bounded intervals (as the endpoints are included) AND bounded functions. The second theorem allows for infinite intervals, e.g. $[1,\infty)$ (improper integral of $\frac1{x^2}$ could be considered there), and also allows for unbounded functions, e.g. $\frac1{\sqrt x}$ when considered on $(0,1]$. In the second theorem $(a,b]$ could conceivably be $(-\infty,b]$, similarly $[a,b)$ could be $[a,\infty)$. – Mirko May 5 '16 at 23:40
• Also, $\sin x/x$ is bounded, and the limit as $x\to 0$ exists, so we can say that $\sin x/x$ is "continuous on $[0,1]$" (abusing language a little bit). What happens is that the definition of improper Riemann integral in fact extends the definition of proper Riemann integral, so we do not have any problem. – Luiz Cordeiro May 5 '16 at 23:42
• yeah but I want to distinguish "Riemann integrable" and "improper Riemann integrable," since the definition of Riemann integrability does not apply to unbounded functions. I think the textbook makes clear of this difference, so that's why I'm confused. – Rainroad May 6 '16 at 5:21

A "proper" Riemann integral works under two assumptions:

1) interval of integration is bounded

2) function being integrated is bounded in that interval.

If any of these assumptions are not applicable then we have to introduce "improper" Riemann integrals as limits of suitable "proper" Riemann Integrals. The first theorem deals with "proper" Riemann integrals where the function is bounded and interval of integration is also bounded.

The second result (in question) deals with the case where interval of integration is bounded, but the function is unbounded in that interval. Note that in this case the function tends to $\infty$ at the end point of the interval. Hence a way to fix this problem is to replace interval $[a, b]$ with $[a + d, b]$ (assuming that function becomes unbounded at the end point $a$) and take limit of integral on $[a + d, b]$ as $d \to 0^{+}$. If the function was unbounded at $b$ then we need to consider the interval $[a, b - d]$ and take limit of integral as $d \to 0^{+}$.

Coming back to your example $(\sin x)/x$, we can see that it is bounded on whole of $\mathbb{R}$, so any integral $$\int_{a}^{b}\frac{\sin x}{x}\,dx$$ exists as a "proper" Riemann integral. The second definition of "improper" integrals does not apply to it. On the other hand we have the standard result $$\frac{\pi}{2} = \int_{0}^{\infty}\frac{\sin x}{x}\,dx = \lim_{x \to \infty}\int_{0}^{x}\frac{\sin t}{t}\,dt$$ so that the integral $\int_{0}^{\infty}((\sin x)/x)\,dx$ exists as an "improper" Riemann integral (this is the case where interval of integration is unbounded, but the function is bounded).

Your textbook probably has a typo when it is trying to introduce the improper integrals. It should say that $f$ is continuous on $(a, b]$ and unbounded on $(a, b]$.

There is another theorem which deals with the case when $f$ is bounded on $(a, b]$ which sort of says that if the function is bounded then taking limits (as in case of improper integrals) does not lead to anything new.

Theorem: If $f$ is bounded on $[a, b]$ and $$\int_{a + d}^{b}f(x)\,dx$$ exists for all $d \in (0, b - a)$ then $$\int_{a}^{b}f(x)\,dx$$ exists as a "proper" Riemann integral. Further $$\lim_{d \to 0^{+}}\int_{a + d}^{b}f(x)\,dx = \int_{a}^{b}f(x)\,dx$$

For a proof of the above theorem see this question. From the above theorem we see that there is no need to introduce the concept of "improper" Riemann integrals when the function and the interval are both bounded.

Note: The use of adjective "proper" here is invented for this answer in order to contrast the Riemann integral with "improper" Riemann integrals.

The Riemann integral is a method to find the area under the curve of a bounded function. The improper Riemann integral is the limit of standard Riemann integrals giving a sensical solution to a problem which cannot be solved by a standard Riemann integral. So what you are really looking at are Riemann integrable functions, and functions that can be estimated arbitrarily well by Riemann integrable functions where that limit exists. So, just because a standard Riemann integrable function can be trivially estimated by other Riemann integrable functions, it doesn't make it improper. An improper integral has to in addition not be defined in a way that allows it to be integrable in a technical sense.

• The textbook makes clear that Riemann integrability only applies to bounded functions, so anything unbounded will be considered non Riemann integrable and only exists as improper (if the limit is finite). The book even says "same definition for unbounded functions," which means they consider both bounded and unbounded continuous functions on (a,b] to be improper, while the first theorem states that bounded functions are Riemann integrable. – Rainroad May 6 '16 at 5:25
• See edit/ overhaul – Nick S May 6 '16 at 16:48