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Vector $\mathbf{y} = (y_1, y_2, ..., y_N)$ and vector $\mathbf{x} = (x_1, x_2, ..., x_N)$ are related according to equation:

$$p(\mathbf{y}) = \frac{1}{|\mathbf{A}|}q(\mathbf{x})$$

And linear equation:

$$\mathbf{y}=\mathbf{A}\mathbf{x}$$

Now I need to change variables for an integral below from $\mathbf{y}$ to $\mathbf{x}$.

$$g(\mathbf{y}) = \int p(\mathbf{y})~ d\mathbf{y}$$

Textbook says its like this:

$$g(\mathbf{y}) = \frac{1}{|A|} \int q(\mathbf{x})~ \bigg| \frac{\partial \mathbf{y}}{\partial \mathbf{x}} \bigg|d\mathbf{x}$$

$$\frac{\partial \mathbf{y}}{\partial \mathbf{x}} = \frac{\partial }{\partial x} Ax = A$$|

$$g(\mathbf{y}) = \frac{1}{|A|} \int q(\mathbf{x})~ d\mathbf{x}$$

I'm a little bit confused by the use of the Jacobian $\bigg| \frac{\partial \mathbf{y}}{\partial \mathbf{x}} \bigg|$ to change variables in this case... mainly because most of the examples on the internet assume you have two integrals when using Jacobian to change variables...

Questions as follows:

(1) How does the Jacobian works in the case of a single integral where the variables x and y are vectors.

(2) I can see that $dy$ in numerator cancels with $\partial{y}$ in denominator... I'm not understanding how the determinate gets removed and why you can cancel a partial differential with a regular differential...

(3) why can't I just use the variable change method they teach in "calculus I" instead of using a Jacobin when applied to a vector integral?

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  • $\begingroup$ What is $A$ in the first equation? I also would be a little bit more explicit about what is a varibale and what depends on what. Because I think you want $x$ to be a function of $y$. Then the totatl derivative reads $dy = \frac{\partial y}{\partial x} dx$. You also don't seem to have a 1-D integral but a N-D one. So your formulas use a bit abuse of notation I think. $\endgroup$ – hal4math Oct 6 at 13:12
  • $\begingroup$ |A| is a scalar constant because its the determinant of a square matrix where all elements are constants... p(y) is a function that takes vector y and returns a vector... g(x) is a function that takes a vector x and returns a vector....all vectors are of length N. I don't know if its abuse of syntax... its just higher level math that you can never figure out which course was suppose to have taught it because non of them actually teach it that way in undergrad classes. I was thinking that maybe a vector integral is really N integrals... not sure about that one... $\endgroup$ – Leaky Capacitor Oct 6 at 13:20
  • $\begingroup$ But notice that $p(y) = \frac{1}{|A|}g(x)$ is an equation where on the left you plug in $y$ and on the right you plug in $x$. You also need an relationship between $x$ and $y$ because otherwise $p$ and $g$ would be constant functions. And why do you think you could use "calculus I" since this is not a 1-D integral but a N-D integral. $\endgroup$ – hal4math Oct 6 at 13:25
  • $\begingroup$ I think its ok to do that... they are just leaving the functions p(~) and g(~) as a mystery...g(x) could be a log function for instances and p(x) could be the log wizbang function .... the point is that you can leave them as mysteries and still solve it.... apparently you need to use the jacobian to do that... I was just curious to understand it better.. $\endgroup$ – Leaky Capacitor Oct 6 at 13:28
  • $\begingroup$ Being precise about what one means is often the key to understanding confusing things. Especially when it comes to short hand notation. Notice for example that $|\frac{\partial x}{\partial y}|$ is quite meaningless in this context: so how is your textbook defining this expression? $\endgroup$ – hal4math Oct 6 at 13:33
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$\newcommand{\R}{\mathbb{R}}$ Let me just summarize the comments and add a bit.

So lets say you have a function $p : \R^N \to \R$ and you would like to calculate the integral of a given (measurable set) $E \subset\R^N$. Then you would write $$ \int_E p(x) dx \qquad \text{which is a real number!} $$ Now, since we are in $\R^N$ this integrale is a $N$-D integral and so $dx$ is really $dx = dx_1dx_2\cdots dx_N$.

Sometimes this can be very difficult to calculate, especially when the set $E$ is a bit unconvenient as it is for example if you want to integrate over a Ball or similar things. Then it makes sense to make a transformation of variables, e.g. polar coordinates. So in general you would have an injective (one-to-one) differentiable function $A$, so that $x = A(y)$. Then $A : \R^N \to A(\R^N)$ is bijective and you can calculate the Jacobian of $A$ which I will denote by $DA$ (here you need a few more extra assumption on this transformaton $A$ to make the following formula work): This is a $N\times N$ matrix in every point $y=(y_1,\dots,y_N)$ and so you can calculate the determinant. Then you get the transformation rule: $$ \int_E p(x) dx = \int_{A^{-1}(E)} p(A(y)) |\text{det}(DA)(y)|dy $$ Now, the expression $|\text{det}(DA)(y)|$ (btw. the $| \cdot |$ mean the modulus here) might be a little bit to long and also there is nice analog from 1-D calculus, so some authors just like to introduce the notation: $$ \left|\frac{\partial y}{\partial x}\right| := |\text{det}(DA)(y)| $$ (I do not like it too much, but that is of course just taste).

Now lets do you special case: We have $$ x = Ay \qquad \text{where} \quad A \quad \text{is a } N\times N \text{matrix}, $$ with non vanishing determinant (so it is bijective!) also we know something about $p(x)$ namely that $$ p(x) = p(Ay) = \frac{1}{|\text{det}(A)|} q(y). $$ What is the first thing we have to do? We need to calculate $DA$. Now, since $A$ is just a constant linear map, the best linear approximation is $A$ itself so we get $DA(p) =A$ for any point $p\in \R^N$.

What is the $E$? Well, it is not mentioned in your setup, so lets just take $E = \R^N$, so $A^{-1}(\R^N) = \R^N$. Nice!

So we get: $$ \int_{\R^N} p(x) dx = \int_{\R^N} p(A(y)) |\text{det}(A)| dy = \int_{\R^N} q(y) dy $$

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Vector Integral Change of Variable Rules

The Jacobian determinant $\bigg|\frac{\partial y}{\partial x} \bigg|$ is needed to change variables of integration that are vectors.

Given:

$$\int_A f(\mathbf{y})~d\mathbf{y}$$

where:

$$\mathbf{y} = g(\mathbf{x})$$

We can change variables of integration from y to x by substitute the Jacobian determinate into the integral as follows::

$$d\mathbf{y} = \bigg|\frac{\partial \mathbf{y}}{\partial \mathbf{x}} \bigg| d\mathbf{x}$$

Then Integrate as following:

$$\int_A f(\mathbf{y})~d\mathbf{y} = \int_{g^{-1}(A)} f(g(x))~\bigg|\frac{\partial y}{\partial x} \bigg| dx$$

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