Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

  2. jcsd
  3. Dec 16, 2013 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

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


    User Avatar
    Science Advisor

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

    should be

    $$\mathrm{d}r = (a^3-A^3)^{1/3} [(\lambda^3-1)^{-1/3}-\lambda^3 (\lambda^3-1)^{-4/3}] \mathrm{d}\lambda$$
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook