Sanjoy Mahajan has written a book called “Street-Fighting Mathematics” that promotes various ways to improve students’ intuition and understanding of topics in applied mathematics. And he starts with dimensional analysis.

A recent Quora question, drawn directly from Mahajan’s book , goes as follows “A common belief is that integration computes areas. Areas have dimensions of L^2. How then can the Gaussian integral have dimensions of L?”

The simple answer here is because the “common belief” is wrong (or at least not always true). But a further important part of the answer is that “the” Gaussian integral Mahajan is referring to is not generic but just a very particular example from his book.

Integration (of real valued functions of a real variable) computes *numbers* which can be related to areas or other things depending on what units are associated with the variables. If integrals computed actual areas then they would give different results depending on the scale at which the graph was plotted – but of course they don’t. What the dimensionless integral of a positive function gives in terms of areas is the *number* of area units between the the graph and the independent axis where one area unit is the area of a rectangle with sides parallel to the axes and spanning one unit on each axis of the graph.

In various applications it makes sense to let variables represent quantities that are not just numbers but have “dimension” which is not in the geometric sense (of 2D vs 3D for example) but refers to the kind of quantity involved (eg Length, Mass, Charge, or whatever). But a probability is just a number so a probability integral doesn’t have any dimension at all. (If its independent variable has dimension D, then its dependent variable is a probability *density* with dimension 1/D.) So in order to answer the question we need to look in Mahajan’s book to find the context in which “the” Gaussian integral is alleged to have dimensions of L.

It turns out that “the” gaussian integral Mahajan is addressing is just the *special case* of \( \int_{-\infty}^{\infty} e^{-\alpha x^2}dx \) where x *has been assigned* to have a dimension of L. (He only does this *after* mentioning that the case with any particular numerical coefficient replacing $latex \alpha$ is dimensionless). Since the exponential is a limit of sums of different powers, the exponent must always be dimensionless (and so $latex \alpha$ must have dimension 1/L^2). This means that the exponential is also dimensionless and so the integral he is looking at has the same dimension as dx, namely L.

**Addendum:** A couple of pages after posing the above question Mahajan goes go on to note that in order for a probability integral to be dimensionless the integrand must include a factor of dimension 1/L – which, if the integral is to remain gaussian, must be a constant multiple of $latex \sqrt{\alpha}$. At this point, in order to find the dimensionless constant multiplier that is needed to give a total probability of 1, he introduces the special case where $latex \alpha$ is omitted (or set equal to the dimensionless constant 1) and asks in a further exercise why this is “ok”. The only possible answer there is that in considering this *new* special case he has dropped the idea that x has dimension L and gone back to the type of integral with which he opened that section of his book.