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

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• Mike400
In summary, the conversation discusses a one-to-one transformation of a 3D volume from (x,y,z) to (t,u,v) and the possibility of the partial derivatives of the Jacobian of transformation not existing at certain points in the domain of integration. It is suggested to use improper integrals in such cases, and an example is given for a mapping where the Jacobian is not defined at certain points. It is also mentioned that for finite number of points, it is possible to have smooth functions that behave similarly. The approach for evaluating integrals in such cases is to split the domain into subdomains.
Mike400
TL;DR 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?

Consider $(x,y,z) = (u^{1/3}, v^{1/3}, w^{1/3})$ for $(u,v,w) \in D = [-1,1]^3$. This mapping is one-to-one, but the jacobian $|(uvw)^{-2/3}|/3$ is not defined on $\{(0,v,w)\} \cup \{(u,0,w)\} \cup \{(u,v,0)\}$. The jacobian is however defined and integrable over the rest of $D$, 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 $$x = f(u) \cos w \sin v \\ y = f(u) \sin w \sin v \\ z = f(u) \cos v$$ for $u \geq 0, v \in [0, \pi], w \in [0, 2\pi)$ for which $$\frac{\partial(x,y,z)}{\partial(u,v,w)} = f^2(u)f'(u) \sin v$$ and choose $f$ so that $f^2(0)f'(0)$ 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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## 1. What is the Jacobian and why is it important in integrals?

The Jacobian is a mathematical concept that represents the rate of change of one set of variables with respect to another set of variables. In integrals, the Jacobian is used to transform the variables in an integral to simplify the calculation process.

## 2. What does it mean if the Jacobian doesn't exist at finite points in the domain of the integral?

If the Jacobian doesn't exist at finite points in the domain of the integral, it means that the transformation used in the integral is not well-defined at those points. This can lead to difficulties in calculating the integral or even render it impossible to solve.

## 3. How does the non-existence of the Jacobian affect the result of the integral?

If the Jacobian doesn't exist at finite points in the domain of the integral, it can lead to incorrect or undefined results. This is because the transformation used in the integral is not valid at those points, and thus the integral cannot be properly evaluated.

## 4. Can the non-existence of the Jacobian be overcome in integrals?

In some cases, it is possible to overcome the non-existence of the Jacobian by using alternative methods or transformations. However, in some cases, it may not be possible to find a valid transformation and the integral may be unsolvable.

## 5. How can one determine if the Jacobian exists at finite points in the domain of the integral?

The existence of the Jacobian can be determined by checking the differentiability of the transformation used in the integral. If the transformation is not differentiable at certain points, then the Jacobian will not exist at those points.

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