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?
$\endgroup$ 132 Answers
$\begingroup$$\newcommand{\R}{\mathbb{R}}$Let me just summarize the comments and add a bit.
So lets say you have a function $p : \R^N \to \R$ and you would like to calculate the integral of a given (measurable set) $E \subset\R^N$. Then you would write$$ \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 $$
$\endgroup$ $\begingroup$Vector Integral Change of Variable Rules
The Jacobian determinant $\bigg|\frac{\partial y}{\partial x} \bigg|$ is needed to change variables of integration that are vectors.
Given:
$$\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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