Is indefinite integration suspect?

In this post, Qiaochu Yuan remarks that 'it is convenient but misleading to write $$ \int f(x) \, dx=g(x) $$ [where the derivative of $g$ is $f$]'. This sentiment seems to be shared by many contributors here, and I don't understand why. To me, both definite and indefinite integration are both valid operations you can perform on a function, and there is nothing suspect about indefinite integration.

I know about the fundamental theorem of calculus, which (as far as I understand) explains the link between indefinite and definite integration. If by integration we mean computing the area under the graph, the fundamental theorem of calculus shows us that integration is the opposite of differentiation, since $$ \frac{d}{dx} \int_{a}^{x} f(t) \, dt = f(x) $$ This shows that every continuous function has an antiderivative. Since a clear link between integration and antidifferentiation has been established, we give the antiderivative the convenient label 'indefinite integral'. (This also explains why the definite and indefinite integration notations are so similar.) This label is fine, so long as we remember that integration is defined as finding the area under the graph, while antidifferentiation is defined as finding the inverse of the derivative.

Another result of the fundamental theorem of calculus is that $$ \int_{a}^{x}f(t) \, dt=\int f(x) \, dx $$ So obviously every indefinite integral can be rewritten in terms of definite integrals, but I don't understand the motivation behind this. If $F$ is an antiderivative of $f$, then why is it more correct to write $$ \int_{a}^{x} f(t) \, dt = F(x) \, , $$ compared to $$ \int f(x) \, dx = F(x) \, ? $$

Solution 1:

Basically, there's a type error: "$\int f(x)\,dx$" is a perfectly meaningful thing, but that thing is not a single function - rather, it's a set of functions.

The point is that a function doesn't have a unique antiderivative. For example, ${x^2\over 2}$ is an antiderivative of $x$ (with respect to $x$ of course), but so is ${x^2\over 2}-4217$. It's not the indefinite integral which is suspect, but rather the notation we use around it - specifically, the way we use "$=$." Properly speaking, $\int f(x)dx$ refers to a set of functions.

This is generally addressed by including a constant of integration, so that we write $$\int x\,dx={x^2\over 2}+C$$ to mean "The set of antiderivatives of $x$ is the set of functions of the form ${x^2\over 2} + C$ for $C\in\mathbb{R}$."

• That said, blindly adding a constant of integration still doesn't always fix the problem: let $f(x)=-{1\over x^2}+1$ if $x>0$ and $-{1\over x^2}-1$ if $x<0$; what's the derivative of $f$, and does $f$ have the form $-{1\over x^2}+C$ for some fixed real number $C$?

Solution 2:

The other answers have made good points about constants of integration but this is not actually what I meant, although it is related. What I meant is what lulu says in the comments: writing antiderivatives this way misleads you about the relationship between the $x$ on the LHS (which is a dummy variable) and the $x$ on the RHS (which is not). The "real" $x$ on the LHS is one of the bounds of integration, which is being suppressed in the notation.

The sense in which this is misleading becomes clearer once you start considering double integrals, which is the context of the question you link to. If it makes sense to write $\int f(x) \, dx = g(x)$, then surely it also makes sense to write $\int g(x) \, dx = h(x)$, right? Then does it make sense to write

$$\iint f(x) \, dx \, dx = h(x)$$

or not? What do you think?

Solution 3:

For one example, the familiar old "formula" $$ \int \frac{1}{x} \ dx = \ln|x| + C $$ is false (unless you define the indefinite integral VERY carefully). This purports to say that any antiderivative of $f(x) = \frac{1}{x}$ must take the form $F(x) = \ln|x|+C$ for some fixed constant $C$. But this is only true over a connected interval. For example, the function $$ G(x) = \begin{cases} \ln|x| +1, & x < 0\\ \ln|x|-1, & x > 0\end{cases} $$ satisfies $G'=f$, even though it is not expressible in the form $\ln|x|+C$. Done right, we should only define indefinite integrals over intervals (this is due to the Mean Value Theorem).

Solution 4:

I think one clear way to see this is that the most pedantic notation would be something like this:

$$ f(x) \in \int f'(x)\,dx $$

You just have to think of $f$ as some element in a vector space, e.g. $f \in C^1(a,b)$, the set of continuous functions with continuous first derivative. So the integral operator is a linear transformation, a map of the kind $\int_a^x: C(a, b) \rightarrow C^1(a, b)$. This makes evident that the writing $$F(x) = \int f(x)\,dx,$$ where $F'(x) = f(x)$, could easily be notation abuse, though most of the time it's not the case, as the others have pointed out. The writer knows what they're writing: they assumed $F$ as a stand-in for a whole set of functions, but this is not always clear to the reader. Another problem this brings to surface is that the integral operator $\int_a^x$ as it has been written above is ill-defined, as it should point one element $f \in C(a,b)$ to one element $g \in C^1(a,b)$, not to a whole set of them. How to define this integral operator soundly is a question above my paygrade.