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HotMintea
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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
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.
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
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.
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