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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

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# Cannot work out change of variables in Integral

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