Riemann integration involving functions that are discontinuous at end points

Question

The definition here says that the function is continuously differentiable on $[a, b]$.

Is it possible to compute the integral when the function is continuously differentiable on $(a, b)$?

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I believe it is possible.

Let us take a function $y$ that is continuous on $[a, b]$ and differentiable on $(a, b) $.

We are to evaluate the integral, $$S=\int_{a}^{b} 2\pi y\sqrt{1+(y')^2} dx$$

Here, the domain of the integrand is $x \in (a, b) $. This tells us that the integrand function is discontinuous at points $a$ and $b$. Since we are computing the Riemann Integral, continuity at end points will not matter.

Even if in another case the integrand was defined at the end points, the areas of the two end rectangles would become insignificant. It would not bother us if the two rectangles were there or not( in the case of discontinuities ). So, the Riemann Integral would give us the same result as it would if the function were undefined at the end points.

In summary, the Riemann integral of end discontinuous functions( not infinite discontinuous ) can be computed.

Answer 1

1. The fact the domain of $f$ is $(a, b)$ rather than $[a, b]$ does not "[guarantee] $f$ is discontinuous at $a$ and $b$," since continuity/discontinuity at a point is only defined for points within the domain of a function, and $a$ and $b$ are not in the domain of $f$. So it just doesn't make sense to talk about continuity/discontinuity at $a$ or $b$.

2. The Riemann integral, defined in terms of partitions of intervals, makes literal sense only on a closed, bounded interval. To handle an open interval, one has to use a limiting process such as $$ \int_{(a,b)} f(x)\, dx = \lim_{a' \to a^{+}} \lim_{b' \to b^{-}} \int_{a'}^{b'} f(x)\, dx. $$

3. If $f$ is continuous on $(a, b)$, it does not follow that $f$ is bounded, i.e., that there exists a real number $M$ such that $|f(x)| \leq M$ for all $x$ in $(a, b)$, i.e., that the graph of $f$ lies between two horizontal lines $y = \pm M$.

   On a closed, bounded interval $[a, b]$, boundedness is automatic from the extreme value theorem.

   Boundedness of $f$ on $(a, b)$ needs to be added as a hypothesis for integrability. For example, $f(x) = 1/x$ is continuously differentiable on $(0, 1)$, but is not (even improperly) integrable on $(0, 1)$. (By contrast, $f(x) = 1/\sqrt{x}$ is not Riemann integrable on $(0, 1)$, but is improperly integrable on $(0, 1)$. This shows boundedness is not necessary for improper integrability, though boundedness is sufficient.)

4. If $f$ is bounded, and is continuous on $(a, b)$, then no matter how $f(a)$ and $f(b)$ are defined (as real numbers), $f$ is integrable on $[a, b]$, and the integral does not depend on the endpoint values.

   This isn't entirely obvious, but your argument sketch furnishes the main idea.

Answer 2

There is no question here, but I'll try to answer anyway.

This is not a matter of beliefs; it is a matter of definitions and of proofs. First of all, the concept of Riemann integral is defined only for bounded functions defined on a closed and bounded interval. So, if your function is not bounded, then it is trivially true that it is not Riemann integrable.

Note that when I wrote that “the concept of Riemann integral is defined only for bounded functions defined on a closed and bounded interval” I was by no means implying that all such function are Riemann integrable. I was only saying that if a function doesn't satisfy these conditions, then it is automatic that it is not is Riemann integrable.

Now let's see a concrete example which is simpler than the one of the area of a surface. The length of the graph of a function $f\colon[a,b]\longrightarrow\mathbb R$ is $\int_a^b\sqrt{1+f'(t)^2}\,dt$. Now consider the function$$\begin{array}{rccc}f\colon&[-1,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\sqrt{1-x^2}\end{array}.$$Since its graph is a semicircle with radius $1$, the length of the graph should be $\pi$. Applying the previous formula, we get that the length of the graph is$$\int_{-1}^1\sqrt{1+f'(t)^2}\,dt=\int_{-1}^1\frac{dt}{\sqrt{1-t^2}}.$$But now we have a problem: the function whose integral we are trying to comput isn't even defined at $\pm1$. Furthermore, it is unbounded. So, it is not Riemann integrable.

What we can do is to work with improper integrals: the previous integrals should be interpreted as$$\left(\lim_{x\to1}\int_0^x\frac{dt}{\sqrt{1-t^2}}\right)+\left(\lim_{x\to-1}\int_x^0\frac{dt}{\sqrt{1-t^2}}\right)\text,$$if both limits exist. And they do! They are both equal to $\frac\pi2$ and so we do get that the length of the graph is equal to $\pi$.