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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?

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up vote 2 down vote accepted

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.

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So, if there's no $\alpha$ - say, just $\int_{a}^{b} e^{x}dx$, then I cannot use dimensional analysis, correct? – andreister Oct 19 '12 at 9:07
Not if $x$ has a dimension. If $x$ is a number (for example a ratio) then you are fine. – Henry Oct 19 '12 at 9:20

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)

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I don't see this. If $x$ is length in metres and $\alpha$ is measured in $m^{-2}$ then the left hand side is dimensionless as is $y$, but the right hand side is not. – Henry Oct 19 '12 at 8:58
The left hand side is not dimensionless as $x$ has units of length... – Fabian Oct 19 '12 at 10:02
OK, I see that now - so if both sides have units of length then $\alpha$ has units of the reciprocal of length squared – Henry Oct 19 '12 at 19:56

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.

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