$f(x) = g(x)$ is equivalent to $\int f(x) \text{dx} = \int g(x) \text{dx}$?

Question

I have a tricky statement. $f(x) = g(x)$ is equivalent to $\int f(x) \text{dx} = \int g(x) \text{dx}$

I don't believe this statement is true. But I cannot decidedly reason why it is not true. My logic goes as follows.

Consider the propositions $A$ and $B$, $f(x) = g(x)$ and $f'(x) = g'(x)$ respectively. Then $A \Rightarrow B$ but $B \nRightarrow A$. $f(x) = 7x + 5$ and $g(x) = 7x + 2$, for example.

There is no problem here. My idea of $\int f(x) \text{dx}$ is that it is any function $F$ such that $F'(x) = f(x)$. If one plugs in $F, G$ respectively into the propositions $A$ and $B$, then one gets that $\int f(x) \text{dx} = \int g(x) \text{dx} \Rightarrow f(x) = g(x)$ but $f(x) = g(x) \nRightarrow \int f(x) \text{dx} = \int g(x) \text{dx}$.

However, I have a problem with this. By substitution under the assumption that $f(x) = g(x)$, the statement $\int f(x) \text{dx} = \int g(x) \text{dx}$ becomes $\int f(x) dx = \int f(x) \text{dx}$. Now clearly this is true, is it not?

I believe the main issue comes from the distinction between letting $\int f(x) \text{dx}$ be a family of all solutions, in which case the sets are the same, or letting $\int f(x) \text{dx}$ be a particular solution $S$ contained within this family of all solutions. My question is, are both distinctions fine to make?

Answer 1

There is two ways to look at this, depending on how you see $\int f(x)\, \mathrm d x$, which kind of an sloppy defined thing. If you see $\int .\,\mathrm d x$ as an operator it is in fact nescessary that if $f = g$ then $\int f\,\mathrm d x = \int g\,\mathrm dx$. This is a basic property we demand of functions.

The thing is: We do have definitions for $\int_a^b f\,\mathrm d x$, including $a,b=\pm\infty$. But the term $\int f\,\mathrm d x$ does not really have a clear definition. Commonly this is used to denote any integral $\int_a^x$ for any lower point $a$ or even any antiderivative of $f$, such as $\int x\,\mathrm dx = \frac{1}{2}x^2 + c$ (note that there are some $c$ such that the right form cannot be expressed as $\int_a^x x \,\mathrm dx$ for any $a$).

By this use you are in fact somewhat correct, in this case $\int$ just becomes a weird way to notate the relation $F$ is antiderivative of $f$ or a placeholder for any such antiderivative. But this is not really the mathematically exact use of integrals, but just a sloppy statement. But your are only somewhat correct, because the inherent problem here is not $f$ and $g$ having different antiderivatives just be changing the name, the problem lies within $f$ alone not having a unique antiderivative. So you could actually even say that $\int f\,\mathrm d x \neq \int f\,\mathrm d x$.

But you can also take equivalence classes by the relation $f$ and $g$ are equivalent if they differ only by constants on each connected subset of our domain $\Omega$. This way you can actually properly define $\int f\,\mathrm dx$ as such an equivalence class, again turning this into a proper operator.

Answer 2

This is primarily a confusion about elementary logic, rather than about calculus specifically. (There are some other problems with your question, such as assuming that $\int f(x) \mathrm{d}x$ is meaningful notation on its own, but the other answers have already addressed those.)

For example, using your reasoning, one could equally well arrive at the following (using your notation): $x = y \Rightarrow x^2 = y^2$, but $x^2 = y^2 \not \Rightarrow x = y$. But substituting $y = x$ results in $x = x$, which is clearly true.

The problem here is (as with $\int$) the sloppy notation. The claim $x = y \Rightarrow x^2 = y^2$ is really the claim: for all real numbers $x, y$, if $x = y$, then $x^2 = y^2$. Because this holds for all reals, you can take $y$ to be any real you want, in particular you can take $y = x$. The claim $x^2 = y^2 \not\Rightarrow x = y$, on the other hand, is really the claim: there are real numbers $x, y$ such that $x^2 = y^2$ but $x = y$. If you write it out explicitly like this, then it is clear why it makes no sense to "substitute" $y = x$.

Your example is very much the same. When you write $\int f \mathrm{d}x = \int g \mathrm{d}x \not \Rightarrow f = g$, what this really means is: there are functions $f, g$ such that $\int f \mathrm{d} x = \int g \mathrm{d} x$ but $f \neq g$. (I am now ignoring the meaninglessness of the notation $\int f \mathrm{d} x$.) Again, clearly it makes no sense to "substitute" $g = f$ here because the claim is not that the equality needs to hold for any choice of $g$ and $f$.

Answer 3

The problem is $\int f(x)\,dx$ is the family or the set of antiderivatives and it only makes sense when the function does have an antiderivative. However Riemann integral and theory of integration is defined in the sense of "definite integrals" or integrals over specific sets. In case of Riemann integral we take compact intervals $[a,b]$ and for Lebesgue we take measurable sets and they give an exact value if the integrals exist(finitely) .

But the notion of $\int f(x)\,dx$ makes sense in terms of solutions of ordinary differential equations with no initial or boundary conditions.

In case of definite integrals this is a much much more subtle issue. The conclusion is false in the case of arbitrary functions even if they are continuous.

For example $\int_{-1}^{1}x\,dx = \int_{-1}^{1} x^{3}\,dx=\int_{-1}^{1}x^{5}\,dx=0=\int_{-1}^{1}0\,dx$ But they are all different as functions.

However they are all different different functions.

However if $f$ is a positive function and $\int_{\Bbb{R}}f\,d\lambda=0$ then it is true that $f=0$ almost surely. That is it is $0$ for all $x$ but for possibly a set of $0$ measure.

For simplicity consider $f$ to be positive and continuous. Then try and prove that $\int_{a}^{b}f(x)\,dx=0$ means $f=0$ in case of Riemann integrals.

Also here is a more general result . If $\int_{A}f\,d\lambda =0$ for all measurable $A$ then $f=0$ almost surely. For Riemann integrals of continuous functions, this is equivalent to this .

BTW I am writing $\int_{A}f\,d\lambda=0\implies f=0$ . But you can replace $f$ by $f-g$ to conlcude about $f=g$.