1.3: Guessing Integrals

The analysis of free fall (Section 1.2) shows the value of not separating dimensioned quantities from their units. However, what if the quantities are dimensionless, such as the 5 and x in the following Gaussian integral:

Alternatively, the dimensions might be unspecified a common case in mathematics because it is a universal language. For example, probability theory uses the Gaussian integral

\[\int_{x_{1}}^{x_{2}} e^{-x^{2}/\sigma^{2}} \label{1.7} \]

where \(x\) could be height, detector error, or much else. Thermal physics uses the similar integral

\[\int_e^{-\frac{1}{2}mv^{2}/kT}dv, \label{1.8} \]

where v is a molecular speed. Mathematics, as the common language, studies their common form \(\int e^{-ax^{2}}\) without specifying the dimensions of \(α\) and \(x\). The lack of specificity gives mathematics its power of abstraction, but it makes using dimensional analysis difficult.

Question

How can dimensional analysis be applied without losing the benefits of mathematical abstraction?

The answer is to find the quantities with unspecified dimensions and then to assign them a consistent set of dimensions. To illustrate the approach, let’s apply it to the general definite Gaussian integral

\[\int_{-\infty}^{\infty} e^{-ax^{2}} dx. \label{1.9} \]

Unlike its specific cousin with \(a = 5\), which is the integral \(\int_{-\infty}^{\infty} e^{-5x^{2} dx,}\) the general form does not specify the dimensions of \(x\) or \(α\) and that openness provides the freedom needed to use the method of dimensional analysis.

The method requires that any equation be dimensionally valid. Thus, in the following equation, the left and right sides must have identical dimensions:

\[\int_{-\infty}^{\infty} e^{-ax^{2}} dx = \text{ something. } \label{1.10} \]

Question

Is the right side \(a\) function of \(x\)? Is it a function of \(α\)? Does it contain a constant of integration?

The left side contains no symbolic quantities other than \(x\) and \(α\). But \(x\) is the integration variable and the integral is over a definite range, so \(x\) disappears upon integration (and no constant of integration appears). Therefore, the right side the “something” is a function only of \(a\). In symbols,

\[\int_{-\infty}^{\infty} e^{-ax^{2}} dx = f(a). \label{1.11} \]

The function \(f\) might include dimensionless numbers such as 2/3 or \(\sqrt{\pi}\), but \(α\) is its only input with dimensions.

For the equation to be dimensionally valid, the integral must have the same dimensions as \(f(α)\), and the dimensions of \(f(α)\) depend on the dimensions of \(α\). Accordingly, the dimensional-analysis procedure has the following three steps:

Step 1. Assign dimensions to \(α\) (Section 1.3.1).

Step 2. Find the dimensions of the integral (Section 1.3.2).

Step 3. Make an \(f(α)\) with those dimensions (Section 1.3.3).

Assigning dimensions to α

The parameter \(α\) appears in an exponent. An exponent specifies how many times to multiply a quantity by itself. For example, here is \(2n\):

\[2^{n} = \underbrace{2 \times 2 \times ... \times 2}. \label{1.12} \]

The notion of “how many times” is a pure number, so an exponent is dimensionless. Hence the exponent −\(αx^{2}\) in the Gaussian integral is dimensionless. For convenience, denote the dimensions of \(α\) by \([α]\) and of \(x\) by \([x]\). Then

\[[α] [x]^{2} = 1, \label{1.13} \]

or

\[[α] = [x]^{-2}. \label{1.14} \]

This conclusion is useful, but continuing to use unspecified but general dimensions requires lots of notation, and the notation risks burying the reasoning.

The simplest alternative is to make \(x\) dimensionless. That choice makes \(a\) and \(f(α)\) dimensionless, so any candidate for \(f(α)\) would be dimensionally valid, making dimensional analysis again useless. The simplest effective alternative is to give \(x\) simple dimensions for example, length. (This choice is natural if you imagine the \(x\) axis lying on the floor.) Then \([α] = L^{-2}\).

Dimensions of the integral

The assignments \([x] = L\) and \([α] = L^{−2}\) determine the dimensions of the Gaussian integral. Here is the integral again:

\[\int_{-\infty}^{\infty} e^{-ax^{2}} dx. \label{1.15} \]

The dimensions of an integral depend on the dimensions of its three pieces: the integral sign \(\int\), the integrand \(e^{−αx^{2}}\), and the differential dx. The integral sign originated as an elongated \(S\) for Summe, the German word for sum. In a valid sum, all terms have identical dimensions: The fundamental principle of dimensions requires that apples be added only to apples. For the same reason, the entire sum has the same dimensions as any term. Thus, the summation sign and therefore the integration sign do not affect dimensions: The integral sign is dimensionless.

Problem 1.6 Integrating velocity

Position is the integral of velocity. However, position and velocity have different dimensions. How is this difference consistent with the conclusion that the integration sign is dimensionless?

Because the integration sign is dimensionless, the dimensions of the integral are the dimensions of the exponential factor \(e^{−αx^{2}}\) multiplied by the dimensions of dx. The exponential, despite its fierce exponent −\(αx^{2}\), is merely several copies of \(e\) multiplied together. Because e is dimensionless, so is \(e^{-ax^{2}}\).

Question

What are the dimensions of dx?

To find the dimensions of dx, follow the advice of Silvanus Thompson [45, p. 1]: Read d as “a little bit of.” Then dx is “a little bit of \(x\).” A little length is still a length, so dx is a length. In general, dx has the same dimensions as \(x\). Equivalently, d the inverse of \(\int\) is dimensionless.

\[\left[\int e^{-\alpha x^{2}} d x\right]=\underbrace{\left[e^{-\alpha x^{2}}\right]}_{1} \times \underbrace{[d x]}_{L}=L. \label{1.16} \]

Problem 1.7 Don't integrals compute areas?

A common belief is that integration computes areas. Areas have dimensions of \(L^{2}\). How then can the Gaussian integral have dimensions of L?

Making an f(α) with correct dimensions

The third and final step in this dimensional analysis is to construct an \(f(α)\) with the same dimensions as the integral. Because the dimensions of \(α\) are \(L^{-2}\), the only way to turn \(α\) into a length is to form \(a^{-1/2}\). Therefore,

\[f(a) ∼ a^{-1/2}. \label{1.17} \]

This useful result, which lacks only \(a\) dimensionless factor, was obtained without any integration.

To determine the dimensionless constant, set \(α = 1\) and evaluate

\[f(1) = \int_{-\infty}^{\infty} e^{-x^{2}}. \label{1.18} \]

This classic integral will be approximated in Section 2.1 and guessed to be \(\sqrt{\pi}\). The two results \(f(1) = \sqrt{\pi}\) and \(f(α) ∼ α^{-1/2}\) require that \(f(a) = \sqrt{\pi/a}\), which yields

\[\int_{-\infty}^{\infty} e^{-ax^{2}}dx = \sqrt {\frac{π}{a}}. \label{1.19} \]

We often memorize the dimensionless constant but forget the power of \(α\). Do not do that. The α factor is usually much more important than the dimensionless constant. Conveniently, the α factor is what dimensional analysis can compute.

Multiple problems

Problem 1.8 Change of variable

Rewind back to page 8 and pretend that you do not know \(f(α)\). Without doing dimensional analysis, show that \(f(α) ∼ α^{-1/2}\).

Problem 1.9 Easy case \(α = 1\)

Setting \(α = 1\), which is an example of easy-cases reasoning (Chapter 2), violates the assumption that \(x\) is a length and \(α\) has dimensions of \(L^{-2}\). Why is it okay to set \(a = 1\)?

Problem 1.10 Integrating a difficult exponential

Use dimensional analysis to investigate \(\int_{0}^{\infty} e^{-at^{3}}\) dt.