Dimensional analysis for integratation.

[HotMintea]

1. The problem statement

Use dimensional analysis to find \int\frac{ dx }{ x^2 + a^2}.

A useful result is \int\frac{ dx} {x^2 + 1}\, \ = \, \ arctan{x} + C.

p.11, prob.1.11 http://mitpress.mit.edu/books/full_pdfs/Street-Fighting_Mathematics.pdf

2. The attempt at a solution

2.1. If I let [x] = L, then [dx] = L and [x^2+1] = L^2. Thus, I expect [\int\ \frac{dx}{x^2+1}]\, \ = \, \frac{1}{L}. However, arctan{x} is dimensionless.

2.2. By the same reasoning as in 2.1., I expect [\int\ \frac{dx}{x^2+a^2}]\, \ = \, \frac{1}{L}. However, there seems to be a multitude of possibilities: \int\ \frac{dx}{x^2+a^2}\, \ = \, \frac{dimensionless\ factor}{x}\, \ , \, \frac{dimensionless\ factor}{a}\ , \frac{dimensionless\ factor}{x\ +\ a}\, \ or\, \frac{a(dimensionless\ factor)}{x^2}, etc.

 

[Dick]

If you pick [x]=L as you have to if [a]=L, then arctan(x) is not dimensionless. It has no particular dimension at all. You'd better pick the argument of arctan to be something other than x. What's a dimensionless argument for arctan?

 

[HotMintea]

What's a dimensionless argument for arctan?

2.1. \int\ \frac{dx}{x^2+1}\ = \frac{arctan{\frac{x}{1}}}{1}\ + \ C, thus [\int\ \frac{dx}{x^2+1}]\ = [\frac{arctan{\frac{x}{1}}}{1}] = \frac{1}{L}.

2.2. \int\ \frac{dx}{x^2+a^2} must cover the case a = 1, thus \int\frac{dx}{x^2+a^2}\ \ = \frac{arctan{\frac{x}{a}}}{a}\ \ + \ C.

Thanks for your help!