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I'm confused about using dimensional analysis, as in Street-Fighting Mathematics, of integrals like $$ \int \! x \cos x \, \textrm{d}x = \textrm{something} $$

  1. I start by expressing the right side as a function of one or more variables. It seems this should be a function of $x$ but it might additionally be a function of $\cos x$, so let's just say $x$ for now $$ \int \! x \cos x \, \textrm{d}x = f(x) $$

  2. I then assign dimensions to $x$ but this is confusing because the first $x$ is used as a scaling factor and the second $x$ is used as an angle (in radians or in degrees, I don't think it matters at this point). Let's just say $x$ is a length in both cases $$ [x] = L $$

  3. I then find the dimensions of the integral but this is also confusing because I'm multiplying a dimension with the cosine of the same dimension. Let's just assume $[\cos x] = L$ $$ \left[\int \! x \cos x \, \textrm{d}x \right] = [x][\cos x][\textrm{d}x] = L^3 $$

  4. I finally make $f(x)$ with the same dimensions as the integral. Because the dimensions of $x$ are $L$, I have $$ f(x) \sim L^3 $$

If I compare this result with the actual antiderivative, I'm really confused because I'm adding different dimensions irrespective of the constant of integration $$ [x\sin x + \cos x + C] = L^2 + L + ? $$

If I compare with the series expansion, then it seems to make sense when $n=1$ $$ [x \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}] = L \times L^2 = L^3 $$

This is all very confusing though so any clarifications would be much appreciated.

Chewers Jingoist
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1 Answers1

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The argument of the cosine is necessarily dimensionless (otherwise the series expansion, which involves adding different powers of $x$, would not work). Therefore $x$ is dimensionless, and thus the integral is dimensionless.

(radians and degrees have dimension of length/length, i.e. they are actually dimensionless)

Wouter
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