I have just read a fascinating proof of the value of the integral $$ \int_{-\infty}^\infty e^{-ax^2} dx, $$ which proceeds by dimensional analysis, as follows: we know that we can write $$ \int_{-\infty}^\infty e^{-ax^2} dx = f(a) $$ for some $f$. Suppose $x$ represents some length, so that $x$ has dimension $[L]$. The argument of the exponential function must be dimensionless, so $a$ must have dimension $[L]^{-2}$. On the LHS, $e^{-ax^2}$ has dimension $[1]$, and $dx$ has dimension $[L]$, so $f(a)$ must also have dimension $[L]$. Hence, we can write $$ f(a) \sim \frac{1}{\sqrt a} $$ where $\sim$ represents proportionality with respect to a dimensionless constant. Now, we need only invoke the well-known result $$ \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}, $$ which shows that $f(1) = \sqrt{\pi}$. Thus, we have $f(a) = \frac{\sqrt{\pi}}{\sqrt a}$, and $$ \int_{-\infty}^\infty e^{-ax^2} dx = \frac{\sqrt{\pi}}{\sqrt a}. $$ This approach of evaluating an integral by dimensional analysis is one that I have never seen before, and it is not obvious to me that I should accept its validity. Why should I expect an equation to remain dimensionally correct when I introduce an arbitrary dimensional constraint (in this case, $x$ having dimension $[L]$)? Under what conditions is such a step valid?

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    $\begingroup$ It is generally true although there are a few exceptions. I remember one in chemistry but I can't put a name on the equation. I know it had to do with quotient of molar concentration, if anyone knows! $\endgroup$ – user88595 Apr 13 '14 at 8:09
  • $\begingroup$ @user88595 Strange. I would not expect there to be an exception in a physical application. Are you sure a dimensional conversion constant wasn't just set to 1 and swallowed up in the equation? $\endgroup$ – David Zhang Apr 13 '14 at 8:18
  • $\begingroup$ Basically the molar concentrations were set to a certain power depending on the reaction so : $\frac{c_1^{d_1}}{c_2^{d_2}}$. Dimensionally it didn't make much sense so the result was dimensionless by convention. $\endgroup$ – user88595 Apr 13 '14 at 8:23
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    $\begingroup$ This dimensional analysis seems to be a rediscovery of the fact that linear substitutions work. Is this impression accurate? $\endgroup$ – anon Apr 13 '14 at 8:45
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    $\begingroup$ @user88595 you are talking about about the equilibrium constant K. Which is defined as concentration of products over reactants. At introductory chemistry, they say K is dimensionless, although mathematically we can easily see this is not the case if we just analyze concentration units. This problem is resolved when you go to physical chemistry and learn about "activities" of solutions. $\endgroup$ – Jeff Faraci Apr 13 '14 at 13:35

It appears with the tip of @anon, I have discovered the answer to my own question.

The justification is simple; namely, linear substitutions work for integrals. Here's the idea: take $$\int_{-\infty}^\infty e^{-ax^2} dx = f(a)$$ and make the substitutions $x \to kx$ and $a \to \frac{a}{k^2}$. This gives $$k \int_{-\infty}^\infty e^{-ax^2} dx = f\left( \frac{a}{k^2} \right).$$ With a bit of rearrangement, we have the functional equation $$f(a) = \frac{1}{k} f\left( \frac{a}{k^2} \right)$$ whose unique solution on $(0, \infty)$ is $$f(a) = \frac{f(1)}{\sqrt a}.$$ This completes the proof without making any use of dimensional constraints, and is easily seen as equivalent to the dimensional analysis method demonstrated above.

  • $\begingroup$ Indeed, in the presence of several units, one has an action of the group $(\mathbb R^+)^n$ (the direct product of $n$ copies of the multiplicative group of real numbers) scaling each of the independent units, and everything should be invariant under this group. $\endgroup$ – Mariano Suárez-Álvarez Apr 13 '14 at 9:14
  • $\begingroup$ The general theory behind this sort of reasoning is invariant theory. $\endgroup$ – Mariano Suárez-Álvarez Apr 13 '14 at 9:15

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