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Cannot work out change of variables in Integral

  1. Dec 16, 2013 #1
    Hi there,

    in a paper the author obtains the integral
    $$\int_{a}^{\infty} \frac {g(\lambda(r))}{r}\mathrm{d}r$$
    which is claimed to be equivalent to
    $$\int_{a/A}^{1} \frac {g(\lambda(r))}{\lambda (\lambda^3-1)}\mathrm{d}\lambda$$
    making use of the relationship (previously physically justified)
    $$r = (a^3-A^3)^{1/3} \lambda (\lambda^3-1)^{-1/3}$$
    where A is a given geometric length, a is a parameter, and the function g(.) is given.
    I cannot work out the substitution. The limits of integration are fine, but when I compute the differential I get
    $$\mathrm{d}r = (a^3-A^3)^{1/3} [(\lambda^3-1)^{-1/3}+\lambda^3 -(\lambda^3-1)^{-4/3}] \mathrm{d}\lambda$$
    so when substituted as expected the factor $$(a^3-A^3)^{1/3}$$ disappers: yet I do not get the desired result.
    Would anybody please be patient enough to point to the error?

    Thanks
     
  2. jcsd
  3. Dec 16, 2013 #2

    SteamKing

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    It looks like you calculated dr/dλ incorrectly. I would check that first.
     
  4. Dec 16, 2013 #3

    mathman

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    $$\mathrm{d}r = (a^3-A^3)^{1/3} [(\lambda^3-1)^{-1/3}+\lambda^3 -(\lambda^3-1)^{-4/3}] \mathrm{d}\lambda$$

    should be

    $$\mathrm{d}r = (a^3-A^3)^{1/3} [(\lambda^3-1)^{-1/3}-\lambda^3 (\lambda^3-1)^{-4/3}] \mathrm{d}\lambda$$
     
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