# Dimensional analysis for integratation.

1. Jul 3, 2011

### 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 [Broken]

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.

Last edited by a moderator: May 5, 2017
2. Jul 3, 2011

### 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?

3. Jul 3, 2011

### HotMintea

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 $\int\frac{dx}{x^2+a^2}\ \ = \frac{arctan{\frac{x}{a}}}{a}\ \ + \ C$.