I'm reading Street-Fighting Mathematics and not sure if I understand integral dimension analysis. The idea is to "guess" integrals without explicit calculation, by just looking at their dimensions.
It's been a good decade since I last touched integrals so please bear with me!
My answer to the problem is that dimension of $\int_{a}^{b}f(x)dx$ depends on the dimension of the multiplication $f(x)*dx$, so while the integral sign is indeed dimensionless, if we have $f(x)$ being "length per second" and $x$ being "second" then the resulting integral will have "length" dimension.
But the author also tells that $e^{-\alpha x^2}$ is dimensionless and thus derives dimension formula for $\alpha$:
I don't understand why the $-\alpha x^2$ should be dimensionless. What if there's no $\alpha$ (or $\alpha=1$), in which case the exponent becomes $e^{-x^2}$ - will it have a dimension now? And if it won't, then why dimension of $\alpha$ should depend on the dimension of $x$ at all?
Also, a final question - if we accept that the only dimension which affects the integral in question is the dimension of $x$, then setting it to "length" means that the integral dimension is "length". But integrals compute areas, don't they? I think that in this case the integral $\int_{a}^{b}f(x)dx$ will have a dimension of "length x 1" which is still area. Is this correct, is there a better explanation?