Is this function well-defined or just an abuse of notation?

Question

Let's say we have a real-valued function $f$, perhaps something very simple. I define $g$ to be $g(x) = \int_0^xf(x)dx$.

This looks like total nonsense to me. But I can't satisfactorily prove to myself that is nonsense. It's clear that $h(x)=\int_0^tf(x)dx$ is well-defined (given some $t$) and it's also clear that $j(x)=\int_0^xf(t)dt$ is well defined. I'm a little rusty on my calculus, but I'm pretty sure that $h(x)$ and $j(x)$ are differentiable for $x\in\Bbb R$.

It seems to me impossible that $g(x)$ should equal either $h(x)$ or $j(x)$, though I imagine people might see $g(x)$ and wonder if it was a typo of either $h(x)$ or $j(x)$. Because just by looking at it, it seems meaningless. But I can't formally explain why.

EDIT: Turns out $h(x)$ is not well-defined. Thanks to those who gave answers.

Answer 1

In your "definition" of $g$ the variable $x$ is used both as an independent variable and as variable of integration. That makes your definition ambiguous, if not nonsensical.

Your "function" $h$ you are also using $x$ as variable of integration. That means, it has no meaning outside of the integral. Therefore your definition of $h$ makes no sense.

The only one that does makes more or less sense is your $j$.

But also here you have to be careful, as $f$ has to be integrable for the integral to make sense. If $f$ is, say, continuous, you are fine and in that case, as you noted, $j$ is even differentiable by the fundamental theorem of calculus.

Answer 2

The discrete analog of this would be a sum like $$\sum_{k=0}^k a_k,$$ which is, of course, nonsense. You can't define the range of values $k$ takes depending on $k$. By the same argument $$\int_0^x f(x)\ \mathrm dx$$ is nonsense.

Even your proposed function $h$ is nonsense, look at the discrete analog $$\widetilde h(k) = \sum_{k=0}^N a_k.$$ You see that $k$ is not a free variable on the RHS where you can plug in some value, it is a bound variable, it's the index used for summation.

Only the function $j$ is well a well defined function. You plug in the upper bound and get a definite integral.

Answer 3

There are two scopes relative to the integral: the outer scope (containing $g(x)$ and $\int_{0}^{x}$) and the inner scope (containing $f(x)$ and $dx$). Because $dx$ must be local to the inner scope, the integration variable must be a different variable than the $x$ in $g(x)$ and $\int_{0}^{x}$. Let's call it $t$. What remains ambiguous is whether the argument of $f$ should be $x$ or the newly renamed $t$. Both make sense: $$ j(x)=\int_{0}^{x}f(t)dt, $$ and $$ k(x)=\int_{0}^{x}f(x)dt=xf(x). $$ The former is more likely to be the intended interpretation.