What does 'dimensions' in this context even mean?

Question

The book on Integral Calculus by Joseph Edwards in its 24th Art. says:

Also it is a further help to observe the dimensions of each side. For instance, $x$ and $a$ being supposed linear, $\int \frac{dx}{\sqrt{a^2 - x^2}}$ is of zero dimensions. There could therefore be no $\frac{1}{a}$ prefixed to the integral.

My question is what does the author mean by dimensions?

Here is a similar question but it doesn't really answer what the term dimension means in this context.

Here is an archive if you want to read it yourself.

Answer 1

The dimensions here are related to what a physicist would think of as units.

Suppose that in this example $x$ represents a length, perhaps measured in centimeters. Then $dx$, which is a small change in $x$, is also measured in centimeters.

In the denominator, $x^2$ has units square centimeters. In order to subtract that from $a^2$ the units must match, so $a$ must be measured in centimeters too. Then $\sqrt{a^2 - x^2}$ in the denominator has units centimeters.

It follows that the quotient in the integrand is unitless (what the author calls dimensionless). Since the integral is just a sum, the definite integral is dimensionless too.

That's why the author can conclude that it must be wrong if you end up with the constant $1/a$ in front of the integral.

Answer 2

Presumably it is meant that the $dx$ portion has dimension $1$, which cancels out the one dimension in the denominator. This coincides with $\frac{dx} {a^2+x^2}$ having $-1$ dimensions. Abusing terminology, $dx$ is usually considered to be a "length," a "difference" in "adjacent" values of $x$. Some mean this more literally than others.