[Math] $f$ is integrable but has no indefinite integral

analysiscalculusintegrationreal-analysis

Let $$f(x)=\cases{0,& $x\ne0$\cr 1, &$x=0.$}$$

Then $f$ is clearly integrable, yet has no antiderivative, on any interval containing $0,$ since any such antiderivative would have a constant value on each side of $0$ and have slope $1$ at $0$—an impossibility.

So does this mean that $f$ has no indefinite integral?

EDIT

My understanding is that the indefinite integral of $f$ is the family of all the antiderivatives of $f,$ and conceptually requires some antiderivative to be defined on the entire domain. Is this correct?

Best Answer

This is a matter of definitions. Usually an antiderivative of a function $f$ is any function whose derivative is $f$. The indefinite integral usually denotes the set of all antiderivatives.

Since every derivative satisfies the intermediate values property, your function $f$ cannot be a derivative and hence has no indefinite integral.

A different thing is the integral function which might be defined as a function $F$ such that $$ \int_a^b f(x) dx = F(b)-F(a). $$ This exists for all integrable functions.

The fundamental theorem of Calculus states that the two concepts agree for continuous functions.

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Edit: Now that your function is bounded, by Lebesgue's criterion for Riemann integrability, your function must not be continuous a.e. Now this link gives you the whole gamut of differentiable functions that have derivatives that are discontinuous on sets of different measures/sizes. https://math.stackexchange.com/a/423279/231021

You want the one in the last part "Increasing the measure". That will guarantee that your function is discontinuous on a set of Lebesgue measure greater than $0$. Now you have to ensure that it is bounded and the example in the comments is one of those.

Here's another link, where they talk about construction of such functions with bounded derivatives everywhere on$[0,1]$ but such that the derivatives are discontinuous on a set of positive measure. (see 5.5.5.d):

https://books.google.co.uk/books?id=1WY6u0C_jEsC&pg=PA212&lpg=PA212&dq=derivative+discontinuous+measure+bounded&source=bl&ots=JtO97c8t60&sig=5YGv2dnZpJ-OO5Uzdodt7BRwnuk&hl=en&sa=X&ei=XvY2VYy5A4fpaKClgbAO&ved=0CEoQ6AEwBg#v=onepage&q=derivative%20discontinuous%20measure%20bounded&f=false

Old post before the bounded restriction: Try unbounded functions. They cannot have Riemann integrals, yet some are continuous on an open interval. What about $x\mapsto \frac{1}{x}: (0,1) \rightarrow \mathbb{R}$?

$\int_0^1 \frac{1}{x} dx$ does not exist (Riemann integrable functions are by definition bounded, and this function is not, on $(0,1)$) but it has an antiderivative $\ln(x)$ everywhere on $(0,1)$.

[Math] Antiderivatives and the indefinite integral

The definite integral $\int_a^b f(x) dx$ can be expressed, via antiderivatives and the fundamental theorem as \begin{equation} \int_a^b f(x) dx = G(b) - G(a). \end{equation} Here you have specified from where to where you want to integrate, making the integral definite. On the other hand, the indefinite integral is the collection of all these integrals, for all possible limit points. So once you make a choice about the integration limits, everything is perfectly well-defined and single-valued, but if you do not do that, you can only determine the (indefinite) integral up to a constant: $$ \int f(x) dx = G + C $$ Since no matter which number $C$ you plug into this equation, the derivative of the right-hand side will always be just $f(x)$. In this way, the indefinite integral is really a collection of functions: $$ \int f(x)dx = \{ G + C ~|~ C \in \mathbb{R}\}. $$ But for the matter of convenience, we leave away the brackets etc and write the integral as if it was a function with an additive, unspecified constant. So I think your misunderstanding lies within the distinction of indefinite (no limits) and definite (with limits) integrals.

Think of it like this:

A definite integral has the form $$ \int_a^b f(x) dx = G(b) - G(a), $$ where $a, b$ are fixed real numbers. The result will be a real number.
Now you allow one of them, say $b$, to be variable. This then gives you an integral FUNCTION of the form $$ I_a (x) = \int_a^x f(t) dt = G(x) - G(a). $$ Two words on notation: I changed the integration variable from $x$ to $t$ to avoid confusion as $x$ now denotes the upper limit of the integral. Furthermore, since the integral function depends on the lower limit $a$ (different $a$'s will give different functions), I stressed this dependence by writing $I_a$ for the function.
Now if you let $a$ remain unspecified, you will get a set of functions which is the indefinite integral. So it is a set as written above, and once we specify the lower bound $a$ and hence the constant $C = -G(a)$ we get one particular function $I_a$. If we then, finally, ask for the value of this function in a point $b$, we get the definite integral from $a$ to $b$ which is a number.

So: a definite integral is a number, an integral function is a function and an indefinite integral is a collection (or set) of functions.

So in the two sources that you provide, the authors each use a different definition for "indefinite integral". The first one uses the definition of step 3 above whereas the second one chose to stop at step 2.

Let me give you an example: Consider the function $$ f(x) = x^2 + 1. $$ An antiderivative is given by $$ G(x) = \frac{1}{3} x^{3} + x $$ but also for any $C \in \mathbb{R}$ by $$ G_{C}(x) = \frac{1}{3} x^{3} + x + C = G(x) + C. $$ Now to evaluate the integral with known bounds is relatively straightforward, you just have to plug the values of one of the antiderivatives into the formula (it doesn't matter which one since you subtract! So the extra constant $C$ will cancel in the evaluation). You could, for example, calculate $$ \int_{3}^{6} f(t) dt = G(6) - G(3) = G(6) - 12 = 75 - 12 = 63. $$ Now let us fix the lower bound, i.e. we let $a = 3$, and vary the upper bound, i.e. $b = x$. Then we will get a function $$ I_{3}(x) := \int_{3}^{x} f(t) dt = G(x) - G(3) = G(x) - 12 = \frac{1}{3} x^{3} + x - 12. $$ As you can see when looking at the right-hand side of the equation, this is a perfectly normal function and you can perform all kinds of calculations with it. This function can be thought of as a special case of the function $G_{C}$ defined above where $C = - 12$. If we now choose to let our integrals start at a different lower bound, say $a = 1$, we will get a related, but different function: $$ I_{1}(x) := \int_{1}^{x} f(t) dt = G(x) - G(1) = G(x) - (\frac{1}{3} + 1) = \frac{1}{3} x^{3} + x - \frac{4}{3}. $$ This corresponds to $G_{\frac{4}{3}}$ in the above notation. So you see that in this way you really get a collection of functions, via the process described in steps 1-3 above, but you can stop this process at any of the steps and it will be clear how to get to the previous/next step.

Hope that helps.

Andre

EDIT

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[Math] Antiderivatives and the indefinite integral

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