# Cannot work out change of variables in Integral

1. Dec 16, 2013

### muzialis

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. Dec 16, 2013

### SteamKing

Staff Emeritus
It looks like you calculated dr/dλ incorrectly. I would check that first.

3. Dec 16, 2013

### mathman

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

should be

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