Dimension analysis of an integral 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?
 A: The point is that the dimensional analysis will break down unless $-\alpha x^2$ is just a number (and so $e^{-\alpha x^2}$ is too), so $\alpha $ must have the dimensions of $x^{-2}$.  
It then does not matter what value $\alpha $ then takes, so long as it has those dimensions. 
In some cases this might be hidden, for example where the exponent uses a unit like radians or decibels, but in those cases the unit is essentially a ratio or a function of a ratio and so is still effectively dimensionless. 
A: You can obtain the results of dimensional analysis using the substitution $y=\sqrt{\alpha} x$, it is easy to see that 
$$\int  e^{-\alpha x^2}\,dx = \frac{1}{\sqrt{\alpha}} \int e^{-y^2}\,dy.$$
As the remaining integral is independent of $\alpha$, we have that the original integral $\propto \alpha^{-1/2}$. (dimensional analysis usually claims additionally that the remaining integral is of order 1)
A: To extend Henry's answer a bit: The exponential function can be written as $$\mathrm e^{-\alpha x^2} = \exp(-\alpha x^2) = \sum_{n=0}^{\infty} \frac{(-\alpha x^2)^n}{n!},$$ so with the same argument on sums as above used to derive the dimension of an integral you see that you must have $[-\alpha x^2] = 1$ to evaluate the exponential. This is true for all(?) non-monomial functions (i.e. all functions that are not of the form $x^n$) with dimension-less coefficients.
