I What if the Jacobian doesn't exist at finite points in domain of integral?

Summary
We use Jacobian when there is change of variables. Is it possible that for a one to one transformation the Jacobian doesn't exist at finite points in the domain of integration ?
Consider a one to one transformation of a ##3##-##D## volume from variable ##(x,y,z)## to ##(t,u,v)##:

##\iiint_V dx\ dy\ dz=\int_{v_1}^{v_2}\int_{u_1}^{u_2}\int_{t_1}^{t_2}
\dfrac{\partial(x,y,z)}{\partial(t,u,v)} dt\ du\ dv##

##(1)## Now for a particular three dimensional volume, is it possible that one or more of the partial derivatives of the Jacobian of transformation doesn't exist at any of the points in the domain of integration?

##(2)## If so, shall we compute the integral using improper integrals?
 

pasmith

Homework Helper
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Consider [itex](x,y,z) = (u^{1/3}, v^{1/3}, w^{1/3})[/itex] for [itex](u,v,w) \in D = [-1,1]^3[/itex]. This mapping is one-to-one, but the jacobian [itex]|(uvw)^{-2/3}|/3[/itex] is not defined on [itex]\{(0,v,w)\} \cup \{(u,0,w)\} \cup \{(u,v,0)\}[/itex]. The jacobian is however defined and integrable over the rest of [itex]D[/itex], a domain consisting of eight disjoint cubes of unit side. The results on each of these can then be added together.

If you want the jacobian to fail to exist at a finite number of points, then I would look at [tex]
x = f(u) \cos w \sin v \\
y = f(u) \sin w \sin v \\
z = f(u) \cos v
[/tex] for [itex]u \geq 0, v \in [0, \pi], w \in [0, 2\pi)[/itex] for which [tex]
\frac{\partial(x,y,z)}{\partial(u,v,w)} = f^2(u)f'(u) \sin v[/tex] and choose [itex]f[/itex] so that [itex]f^2(0)f'(0)[/itex] is not defined. One can envisage smooth functions which at a finite number of points behave locally in this fashion, although it may not be possible to give formulae for them. So far as actually evaluating integrals over such regions, I would split the domain into subdomains which each contain only a single coordinate singularity, which must correspond to a boundary of the subdomain in (u,v,w) space.
 
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