Does an integral without an integrand imply the integrand is $1$?

Question

This is a question concerning mathematical convention. If we see something like

$$ \int_a^b dx $$

Does this imply that

$$ \int_a^b dx = \int_a^b 1\ dx\text{?} $$

Answer 1

Yes, $1$ is implied in $\int_a^b\, dx = \int_a^b 1\, dx$, similar to how people write $$\int_a^b \frac{\,dx}{x^2} \quad \text{instead of }\quad \int_a^b \frac{1}{x^2}\,dx$$

Notation has its critics, as pretty much every notation for integrals. I vaguely remember a book suggesting $\int_a^b 1(x)\, dx$, because the double appearance of $x$ should remind the reader that it's a dummy variable, like an index of summation. I'm not expecting this to catch on, though.

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At least nobody is writing $\displaystyle \sum_{i=a}^b$ instead of $\displaystyle \sum_{i=a}^b 1$.