1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dimensional analysis for integratation.

  1. Jul 3, 2011 #1
    1. The problem statement

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

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

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

    2.2. By the same reasoning as in 2.1., I expect [itex] [\int\ \frac{dx}{x^2+a^2}]\, \ = \, \frac{1}{L} [/itex]. However, there seems to be a multitude of possibilities: [itex] \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} [/itex], etc.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 3, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  4. Jul 3, 2011 #3
    2.1. [itex] \int\ \frac{dx}{x^2+1}\ = \frac{arctan{\frac{x}{1}}}{1}\ + \ C[/itex], thus [itex] [\int\ \frac{dx}{x^2+1}]\ = [\frac{arctan{\frac{x}{1}}}{1}] = \frac{1}{L}[/itex].

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

    Thanks for your help! :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook