6.9 Jacobian of a composition and improper double integrals
When trying to perform a change of variables in a double integral sometimes one encounters difficulties in computing the Jacobian.
One idea is to decompose the function that gives the change of coordinates as a composition of more simple functions.
Theorem
Let \(D, E \subseteq \mathbb{R}^2\) be two open sets. Let \(\varphi: D \rightarrow E\) and \(\psi: E \rightarrow \mathbb{R}^2\) be two differentiable functions and define \(u=\psi \circ \varphi: D \rightarrow \mathbb{R}^2\). Then
Proof
Let \(\varphi:D\to E\), \(\psi:E\to \R^2\) where \((s,t)\in D\), \((a,b)\in E\) and \((x,y)\in \R^2\)
\(\varphi(s, t)=(a(s, t), b(s, t)), \quad \psi(a, b)=(x(a, b), y(a, b))\)
\(u(s, t)=(\psi \circ \varphi)(s, t)=\psi(\varphi(s, t))=(x(\varphi(s, t)), y(\varphi(s, t)))\)
\(=(x(a(s, t), b(s, t)), y(a(s, t), b(s, t)))=(x\circ \varphi(s,t),y\circ \varphi(s,t))\)
Then \(JM_u=\begin{pmatrix} \frac{\partial(x\circ \varphi)}{\partial s}&\frac{\partial(x\circ \varphi)}{\partial t}\\ \frac{\partial(x\circ \varphi)}{\partial s} & \frac{\partial(y\circ \varphi)}{\partial t} \end{pmatrix}\), then \(\frac{\partial(x\circ\varphi)}{\partial s}\) \(=\nabla x\left(\varphi(s,t)\right)\cdot\left(\frac{\partial a}{\partial s},\frac{\partial b}{\partial s}\right)=\frac{\partial x}{\partial a}\left(\varphi(s,t)\right)\cdot \frac{\partial a}{\partial s}(s,t)+\frac{\partial x}{\partial b}\left(\varphi(s,t )\right)\cdot\frac{\partial b}{\partial s}(s,t)\) we use this
\(\frac{\partial(x\circ\varphi)}{\partial t}=\nabla x(\varphi(s,t))\cdot\left(\frac{\partial a}{\partial t},\frac{\partial b}{\partial t}\right)=\frac{\partial x}{\partial a} (\varphi(s,t))\frac{\partial a}{\partial t}+\frac{\partial x}{\partial b}(\varphi (s,t))\frac{\partial b}{\partial t}\)
\(\frac{\partial(y\circ\varphi)}{\partial s}=\nabla y(\varphi(s,t))\cdot\left(\frac{\partial a}{\partial s},\frac{\partial b}{\partial s}\right)=\frac{\partial y}{\partial a} (\varphi(s,t))\frac{\partial a}{\partial s}+\frac{\partial y}{\partial b}(\varphi (s,t))\frac{\partial b}{\partial s}\)
\(\frac{\partial(y\circ\varphi)}{\partial t}=\nabla y(\varphi(s,t))\cdot\left(\frac{\partial a}{\partial t},\frac{\partial b}{\partial t}\right)=\frac{\partial y}{\partial a} (\varphi(s,t))\frac{\partial a}{\partial t}+\frac{\partial y}{\partial b}(\varphi (s,t))\frac{\partial b}{\partial t}\)
\(\Rightarrow JM_{n}= \begin{bmatrix} \frac{\partial(x\circ\varphi)}{\partial s} & \frac{\partial(x\circ\varphi)}{\partial t} \\ \frac{\partial(y\circ\varphi)}{\partial s} & \frac{\partial(y\circ\varphi)}{\partial t} \end{bmatrix}= \begin{bmatrix} \frac{\partial x}{\partial a}(\varphi(s,t)) & \frac{\partial x}{\partial b}(\varphi(s,t)) \\ \frac{\partial y}{\partial a}(\varphi(s,t)) & \frac{\partial y}{\partial b}(\varphi(s,t)) \end{bmatrix} \begin{bmatrix} \frac{\partial a}{\partial s} & \frac{\partial a}{\partial t} \\ \frac{\partial b}{\partial s} & \frac{\partial b}{\partial t} \end{bmatrix}\)
\(= J M_\psi(\varphi(s,t)) \cdot J M_{\varphi(s,t)}\)
Corollary
Let \(D,S \subseteq \mathbb{R}^2\) be two open sets. Let \(u: D \to S\) be a bijective function with \(u(s,t)=(x(s,t), y(s,t))\) and \(x,y: D \to \mathbb{R}\) be differentiable functions. Assume further that the inverse function \(u^{-1}: S \to D\) has differentiable coordinate functions. Then
In particular, if \(J_{u^{-1}}(u(s,t)) \ne 0\) then
Example
The formula above allow us to compute the Jacobian of a function by computing the Jacobian of the inverse.
Consider the set in \(\mathbb{R}^2\) given by \(S=\{(x,y)\in\mathbb{R}^2 | 1 \le xy \le 3, \quad 3 \le y^2-x^2 \le 4\}\)
Example: Compute the area of
\(S = \{(x,y) \in \mathbb{R}^2 \mid 1 \le \underbrace{xy}_{s} \le 3, \quad 3 \le \underbrace{y^2-x^2}_{t} \le 4 \}\)
Since the set is symmetric, we compute the area for \(\boxed{x,y>0}\) and then multiply by 2.
\(1 \le s \le 3, \quad 3 \le t \le 4\)
\(u(x,y) = (s(x,y), t(x,y))\) \(= (xy, y^2-x^2)\)
\(a(s_{1})=\int_{S_1}1=\int_{D}1\cdot|J_{u^{-1}}|=\int_{1}^{3}\int_{3}^{4}\frac{1}{\sqrt{t^{2}+4s^{2}}} dtds\)
\(J_{u}(x,y) = \det \begin{pmatrix} \frac{\partial s}{\partial x} & \frac{\partial s}{\partial y} \\ \frac{\partial t}{\partial x} & \frac{\partial t}{\partial y} \end{pmatrix} = \det \begin{pmatrix} y & x \\ -2x & 2y \end{pmatrix} = 2y^{2}+2x^{2} = 2(x^{2}+y^{2})\)
\(s = xy\) and \(t = y^2-x^2\) \(= \frac{1}{2\sqrt{t^2+4s^2}}\) since \(t^2 = y^4+x^4-2x^2y^2\) and \(\begin{cases} t^{2}+4s^{2} = y^{4}+x^{4}-2x^{2}y^{2}+4x^{2}y^{2} \\ = y^{4}+x^{4}+2x^{2}y^{2} = (x^{2}+y^{2})^{2} \end{cases}\)
Then \(x^{2}+y^{2} = \sqrt{t^{2}+4s^{2}}\), then \(J_{u^{-1}}(u(x,y))=J_{u^{-1}}(s,t)=\frac{1}{2(x^{2}+y^{2})}=\frac{1}{2\sqrt{t^{2}+4s^{2}}}\)
Improper double integrals
Up now we were computing integrals for functions of two variables on set that satisfy the following:
- The sets \(S \subseteq \mathbb{R}^2\) are \(\underline{\text{bounded}}\) and have area.
- The functions are \(\underline{\text{bounded}}\) and are integrable over \(S\).
In this last section we extend the definition to \({\text{unbounded sets}}\) and/or \({\text{unbounded functions}}\).
We start by considering \({\text{unbounded sets}}\); actually only the whole plane \(\mathbb{R}^2\).
Let us first recall how we define the Improper integral in the case of functions of one variable.
Given a set \([a, b]\) -- where \(b\) could be \(+\infty\) --, a function \(f: [a, b] \to \mathbb{R}^2\) is called integrable on \([a, b]\) if it is integrable on \([a, c]\) for all \(a \le c < b\) and the following limit exists \(I = \lim_{c \to b^-} \int_a^c f(x)dx\)
In such a case, we write \(I = \int_a^b f(x)dx\) and calle it the improper integral of \(f\) on \([a, b]\).
Here in the one-dimensional case, we have only one way to go to \(+\infty\) (or \(-\infty\))
Question:
How do we extend this notions for sets in \(\mathbb{R}^2\)?
Note that we have infinite directions to approach to infinity. 
Definition
Let \(f: \mathbb{R}^2 \rightarrow \mathbb{R}\) be a function. Assume that \(\forall\) bounded set \(S \subseteq \mathbb{R}^2\) with area, \(f\) is integrable on \(S\).
We say that \(f\) is integrable on \(\mathbb{R}^2\) if there is \(I \in \mathbb{R}\) such that \(\forall \varepsilon > 0\), \(\exists r > 0\) s.t. \(\forall\) bounded set \(S\) with area with \([-r,r] \times [-r,r] \subseteq S\), we have \(\displaystyle \left| \int_{S} f - I \right| < \varepsilon\)
