Why is an equation necessarily dimensionally correct?

Question

I have just read a fascinating proof of the value of the integral $$ \int_{-\infty}^\infty e^{-ax^2} dx, $$ which proceeds by dimensional analysis, as follows: we know that we can write $$ \int_{-\infty}^\infty e^{-ax^2} dx = f(a) $$ for some $f$. Suppose $x$ represents some length, so that $x$ has dimension $[L]$. The argument of the exponential function must be dimensionless, so $a$ must have dimension $[L]^{-2}$. On the LHS, $e^{-ax^2}$ has dimension $[1]$, and $dx$ has dimension $[L]$, so $f(a)$ must also have dimension $[L]$. Hence, we can write $$ f(a) \sim \frac{1}{\sqrt a} $$ where $\sim$ represents proportionality with respect to a dimensionless constant. Now, we need only invoke the well-known result $$ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}, $$ which shows that $f(1) = \sqrt{\pi}$. Thus, we have $f(a) = \frac{\sqrt{\pi}}{\sqrt a}$, and $$ \int_{-\infty}^\infty e^{-ax^2} dx = \frac{\sqrt{\pi}}{\sqrt a}. $$ This approach of evaluating an integral by dimensional analysis is one that I have never seen before, and it is not obvious to me that I should accept its validity. Why should I expect an equation to remain dimensionally correct when I introduce an arbitrary dimensional constraint (in this case, $x$ having dimension $[L]$)? Under what conditions is such a step valid?

Answer 1

It appears with the tip of @anon, I have discovered the answer to my own question.

The justification is simple; namely, linear substitutions work for integrals. Here's the idea: take $$\int_{-\infty}^\infty e^{-ax^2} dx = f(a)$$ and make the substitutions $x \to kx$ and $a \to \frac{a}{k^2}$. This gives $$k \int_{-\infty}^\infty e^{-ax^2} dx = f\left( \frac{a}{k^2} \right).$$ With a bit of rearrangement, we have the functional equation $$f(a) = \frac{1}{k} f\left( \frac{a}{k^2} \right)$$ whose unique solution on $(0, \infty)$ is $$f(a) = \frac{f(1)}{\sqrt a}.$$ This completes the proof without making any use of dimensional constraints, and is easily seen as equivalent to the dimensional analysis method demonstrated above.