How one defines $\int_0^t dt$?

Question

I saw in a physics book the computation $\int_0^tdt=t$. But I was wondering how one defines rigorously the integral. I know already how to define $\int_0^a dt$ but I have never seen a rigorous definition of $\int_0^{f(t)}dt$.

Answer 1

In the physics book you read, the expression $$ \int_0^t f(t) dt $$ is meant to represent $$ \int_0^tf(s)ds. $$

Writing $\int_0^t dt = t$ is an abuse of notation: the symbol $t$ was used twice with different meanings and does not represent the same variable in the two places it appears.

Answer 2

Just to be clear, $\int f(x) dx$ is the antiderivative or indefinite integral of $f(x)$ while $\int_{a}^{b} f(x)dx$ is the definite integral and while they are closely related as I will show, they are technically two different things.

Normally, as in the comments, we cannot have bounds of integration also be the variable we are integrating with respect to. But if we "separate" the $t$s into $t$ for the bound and $T$ for the variable being integration, we can see what they are trying to show is that the antiderivative of $1$ with respect to $t$ is $t$.

$$\int f(x)dx=\int_{0}^{x}f(t)dt$$

If and only if the constant of integration C is assumed to be 0

So, in your case:

$$\int dt = \int 1 \cdot dt = \int_{0}^{t}dT=\int_{0}^{t} 1 \cdot dT= t$$

I will say this is very sloppy and confusing notation, and goes against the convention, but can in a way be more convenient if you interpret future integrals set up in this manner like this:

$$\int_{0}^{t}f(t)dt = \int f(t)dt$$