Evaluate $\int_0^1 \int_0^{1-y} \cos \left(\frac{x-y}{x+y}\right)dxdy$

Question

Evaluate the following double integral $$\int_0^1 \int_0^{1-y} \cos \left(\frac{x-y}{x+y}\right)dxdy$$

I tried transforming to \begin{align} x+y &=u\\ x-y &=v \end{align} but I think it is getting complicated. Thanks in advance.

Answer 1

You are on the right track. By letting $u=x+y$ , $v=x-y$, we have $$\left|\frac{\partial (x,y)}{\partial (u,v)}\right|=\left|\frac{\partial (u,v)}{\partial (x,y)}\right|^{-1}=\left|\det \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\right|^{-1}=|-2|^{-1}=\frac{1}{2}.$$ Moreover the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$ in the $xy$-plane is transformed into the triangle with vertices $(0,0)$, $(1,1)$ and $(1,-1)$ in the $uv$-plane. Therefore $$\int_{y=0}^1\left( \int_{x=0}^{1-y} \cos\left(\frac{x-y}{x+y}\right)dx\right)dy=\frac{1}{2}\int_{u=0}^1\left(\int_{v=-u}^u\cos(v/u)dv\right) du.$$ Can you take it from here?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{1}\int_{0}^{1 - y}\cos\pars{x - y \over x + y}\,\dd x\,\dd y \,\,\,\stackrel{x\ \mapsto\ \pars{1 - y} - x}{=}\,\,\, \int_{0}^{1}\int_{0}^{1 - y}\cos\pars{1 - x - 2y \over 1 - y - x + y} \,\dd x\,\dd y \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1 - y}\cos\pars{1 - {2y \over 1 - x}}\,\dd x\,\dd y = \int_{0}^{1}\int_{0}^{1 - x}\cos\pars{1 - {2y \over 1 - x}}\,\dd y\,\dd x \\[5mm] \stackrel{x\ \mapsto\ 1 - x}{=}\,\,\, & \int_{0}^{1}\int_{0}^{x}\cos\pars{1 - {2y \over x}}\,\dd y\,\dd x = \int_{0}^{1}\bracks{-\,{1 \over 2}\,x\,\sin\pars{1 - {2y \over x}}} _{\ y\ =\ 0}^{\ y\ =\ x}\,\,\,\,\dd x \\[5mm] = &\ -\,{1 \over 2}\int_{0}^{1}x\bracks{\sin\pars{-1} - \sin\pars{1}}\,\dd x = \sin\pars{1}\int_{0}^{1}x\,\dd x = \bbx{{1 \over 2}\,\sin\pars{1}} \approx 0.4207 \end{align}