- #1
muzialis
- 156
- 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
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