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Homework Help: 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


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    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:
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