Evaluate $\int_A x^{-1}$

Question

We are asked to evaluate $\displaystyle \int_A x^{-1}dV(A)$, with $A=\{ (x,y):2<x+y<4,y>0,x-y>0\}$. From the solutions we know that$\displaystyle \int_A x^{-1}dV(A)=2Log(2)$. The point is to use the change of coordinates theorem to evaluate the integral.

After proving that $\mathbf{g}(s,t)=(s^2+t^2,s^2-t^2)$ is regular (it's univalent only in each quadrant), I've tried computing $\displaystyle \int_B (s^2+t^2)^{-1}\cdot |det(D\mathbf{g}(s,t))|\text{ }dV(A)$, with $B=\{ (s,t):2<s^2<4,\text{ }s^2>t^2,\text{ and } |s|>|t| \}$.

Well, when trying to compute $\displaystyle \int_B (s^2+t^2)^{-1}\cdot |det(D\mathbf{g}(s,t))|\text{ }dV(A)$,

$\displaystyle \int_B (s^2+t^2)^{-1}\cdot |det(D\mathbf{g}(s,t))|\text{ }dV(A)= \int^2_\sqrt{2}\int^s_0\frac{8st}{s^2+t^2}+\int^2_\sqrt{2}\int^0_{-s}\frac{-8st}{s^2+t^2}+\int^{-\sqrt{2}}_{-2}\int^0_s\frac{8st}{s^2+t^2}+\int^{-\sqrt{2}}_{-2}\int^{-s}_0\frac{-8st}{s^2+t^2}=4Log(16)$

Where did I go wrong? Any help will be appreciated.

Answer 1

Method A: With Change of Variable as chosen by the OP

The OP changed the region A into B with the following variables: $$s^2+t^2=x\ \text{and} \ s^2-t^2=y$$ These equations define a transformation $T^{-1}$ from the $x\text{-}y$ plane to the $s\text{-}t$ plane. The Jacobian of T is:

$$\begin{vmatrix} 2s & 2t \\[0.3em] 2s & -2t \end{vmatrix}=-8st$$ To find the region A in the $s\text{-}t$ plane we look at the sides of the image in Method B.

$$y=0,\ \ x=y,\ \ x+y=4,\ \ x+y=2$$ This transforms to the image lines of B $$s^2=t^2, \ \ t=0, \ \ s^2=1, \ \ s^2=\sqrt{2} $$ Notice that this mapping is not $1\text{-}1$, hence its not possible to directly use the change of variable theorem. If we limit the region to only $s,t>0$ then we would have a one-to-one mapping. So an added constraint is $s,t>0$

The region B should be divided into symmetric parts $B1$ and $B2$ as shown in the image below. However, to insure $1\text{-}1$ mapping the relevant region is only $B1$

Below is the integral to find $B1$

$$\int \limits_{B1} \frac{|-8st|}{s^2+t^2} dtds=\int_{1}^{\sqrt{2}} \int_0^{s} \frac{|-8st|}{s^2+t^2} dtds=\ln(4)$$

Method B: Without Change of Variable

The first thing to do in this exercise is draw the region A. Its a very easy region to draw. I graphed it and the image is shown below,

We are trying to find the double integral over the enclosed region. It is clear that this region needs to be divided into two parts, $A_1$ and $A_2$ as shown in the image below with the red region and green region representing $A_1$ and $A_2$ respectively.

The two integrals are then supposed to be expressed as such:

$$\underbrace{\int_1^2\int_{2-x}^{x}x^{-1}dydx}_{A_1}+\underbrace{\int_2^4\int_{0}^{4-x} x^{-1} dydx}_{A_2}$$ $$\left[2-\ln(4) \right] +\left[\ln(16)-2\right]=\ln(4)=2\ln(2)$$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{? \equiv \int_{A}{\dd V \over x}\quad\mbox{where}\quad A \equiv \braces{\pars{x,y}\quad \ni\quad 2 < x + y < 4,\quad y>0,\quad x - y > 0}}$

The use of the $\it Heaviside\ Step$ function $\Theta:{\mathbb R}\verb*\*\braces{0} \to {\mathbb R}$: $$ \Theta\pars{x} \equiv \left\lbrace \begin{array}{lcl} 0 & \mbox{if} & x < 0 \\[1mm] 1 & \mbox{if} & x > 0 \end{array}\right. $$ avoids any pictures.

\begin{align} \color{#0000ff}{\Large ?}& = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Theta\pars{x + y - 2}\Theta\pars{4 - x - y}\Theta\pars{y}\Theta\pars{x - y}\, {\dd x\,\dd y \over x} \\[3mm]&= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Theta\pars{y - 2}\Theta\pars{4 - y}\Theta\pars{y - x}\Theta\pars{2x - y}\, {\dd x\,\dd y \over x} \\[3mm]&= \int_{-\infty}^{\infty}\dd y\,\Theta\pars{y - 2}\Theta\pars{4 - y} \int_{y/2}^{y}\Theta\pars{y}\,{\dd x \over x} = \int_{2}^{4}\dd y\,\ln\pars{y \over y/2} = \color{#0000ff}{\large 2\ln\pars{2}} \end{align}