Indefinite Integral of all even function $= 0$?

Question

I am wondering what's wrong with the following decent-looking proof: \begin{align*} \int x^2 dx &= \int (-u)^2(-du)\quad\text{ Let } x=-u\\ &=-\int u^2 du\\ &=-\int x^2 dx \quad \text{Change of dummy variable} \\ 2\int x^2 dx&=0\\ \therefore \quad \int x^2 dx&=0 \end{align*} Which is obviously wrong as $\int x^2 dx=\frac{x^3}{3}+C$. But I am not quite sure which part went wrong, great thanks for your help.

Answer 1

This is actually a very good question, and the underlying problem is the difference between definite and indefinite integrals, and the lack of formality when working with definite integrals. Also, and worth noting, the problem doesn't have anything to do with indefinite integrals being defined up to a constant.

If you tried to do the same trick with definite integrals, it wouldn't work.

The integral you wrote is defined as $$ F(x) = \int x^2 dx $$ which formally means $F(x)$ is a function (defined up to some constant) and $F'(x) = x^2$.

The werid thing here is that the variable $x$ is both inside and outside the integral, which doesn't really make sense with definite integrals, but it does with indefinite.

If you do the change of variables now, you get

$$ \int u^2 du = F(u) = F(-x) = \int (-x)^2 d(-x) = -\int x^2dx = -F(x). $$ So the only information you gain is $F(u)=F(-x)=-F(x)$.

Variable change does not work with functions. You cannot assume $F(u)=F(x)$ by just changing $u$ with $x$. And this is exactly what is being done by changing "the dummy variable" inside the integral. You are not actually changing a varibale inside an integral, but rather you are changing the variable of a function.
Therefore, saying $\int a(x)dx = \int a(y)dy$ is wrong, because these are not indefinite integrals, but functions in $x$, resp. $y$.

Answer 2

There is a problem in the “Change of dummy variable” step. The symbol $u$ is not a dummy variable; it stands for $-x$ (and, yes, $x$ is a dummy variable). When you get to $-\int u^2\,\mathrm du$, if $\int u^2\,\mathrm du$ is some function $f$ depending on the variable $u$, then you have obtained $-f(u)$. But now, since $u=-x$, this is equal to $-f(-x)$. So, if $f$ is a primitive of $x^2$, all that this proves is that this primitive an odd function (or to be more precise, the sum of an odd function with a constant; always keep in mind that the expression $\int\varphi(x)\,\mathrm dx$ doesn't stand for a single primitive).