Are the definite and indefinite integrals actually two different things? Where is the flaw in my understanding?

Some context: I'm an engineer, and I tend to have a rather unusual way of understanding and thinking about things, most likely related to my being autistic. I found this question on the HNQ and upon reading it I felt rather confused.

I have always understood that the indefinite and definite integrals are two sides of the same underlying concept, that there is no meaningful difference between them other than that one evaluates to a function and the other to a quantity. Essentially, the definite integral is what you get from running the numbers on the result of an indefinite integral, and saying it's unrelated to the definite integral is like saying that $\sin(x)$ is fundamentally different from $\sin(a)$, assuming $x$ is a variable and $a$ a constant, for no reason other than that $x$ is a variable and $a$ a constant.

To me, whether the definite and indefinite integrals are two sides of the same thing, or two closely related but different things, seems more like a question of philosophy than mathematics.

Where is the flaw in my understanding? Is there one?

Solution 1:

The point is, there are three slightly different concepts here:

1) If $f$ and $F$ are functions with the same domain, and $F'(x)=f(x)$ on that domain, then we say that $F$ is an antiderivative of $f$.

2) If $f$ is a function on the interval $a \le x \le b$, then the definite integral of $f$, $\int_a^b f(x) \, dx$, is (loosely speaking) the area under the graph of $y=f(x)$ for $a \le x \le b$, or (more precisely speaking) a limit of Riemann sums. This definition does not involve any derivatives.

3) If $f$ is a function on the interval $a < x < b$ and $c$ is a point in that interval, then the function $G(x)=\int_c^x f(t) \, dt$ could be called an indefinite integral of $f$. This definition of the function $G$ comes directly from the definition in 2) of the definite integral, so it also does not involve any derivatives.

You are saying that you don't see any difference between 2) and 3). You are correct! If your mental definition of "indefinite integral" is 3), then an indefinite integral is just a definite integral with some unknown limits.

The question you're linking is saying that many calculus books use 1) as the definition of an "indefinite integral": that is, they say that if $F$ is an antiderivative of $f$, then $\int f(x) \, dx = F(x)+C$ is the indefinite integral of $f$. They can get away with this because of the fundamental theorem of calculus, which says (in part) that

$$ \frac{d}{dx}\int_c^x f(t) \, dt=f(x) \, . $$

That is, the function $G(x)$ defined in 3) is an antiderivative of the function $f$, so these two notions of "indefinite integral" are closely related.

But closely related does not mean identical! K B Dave's answer is giving you some examples of situations where the two definitions 1) and 3) diverge slightly. Specifically:

• We gave the definition 3) on an interval $a < x < b$, but it doesn't work very well on a more complicated domain. This means that definition 1) is more robust when $f$ has singularities.
• The fundamental theorem of calculus says that every "indefinite integral" in the sense of 3) is an antiderivative. It does not say that every antiderivative is an indefinite integral, which in fact is false; you can have functions $f$ and $F$ where $F'(x)=f(x)$, but there is no way to write $F(x)=\int_c^x f(t) \, dt$.

Solution 2:

$\DeclareMathOperator{\dom}{dom}$

"Reasonable" functions $f$ can have antiderivatives $D^{-1}f$ that are not of the form $x\mapsto \int_{a}^xf(t)\mathrm{d}t$ for any $a$ in the domain of $f$.

1. Let $U=(-\infty,0)\cup(0,\infty)$, and consider $$\begin{align} f&: U\to\mathbb{R} & x&\mapsto \frac{1}{x}\text{.} \end{align}$$ Then $$D^{-1}f(x)=\begin{cases} \ln(-x)+C_- & x < 0\\ \ln(x) + C_+ & x > 0 \end{cases}$$ where $C_-$ and $C_+$ can differ—because the domain is not path-connected, there is no way to integrate from positive $x$ to negative $x$.
2. Consider $$\begin{align} f&:\mathbb{R}\to \mathbb{R} & x&\mapsto \frac{1}{1+x^2}\text{.} \end{align}$$ Then $$\int_a^xf(t)\mathrm{d}t=\arctan x - \arctan a$$ but $$D^{-1}f(x)=\arctan x + C\text{;}$$ since the magnitude of $\arctan a$ is bounded by $\pi/2$, if $\lvert C \rvert \geq \pi/2$ then $D^{-1}f(x)$ is not of the form $\int_a^xf(t)\mathrm{d}t$.