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

How can I evaluate the following;

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

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

• For extended point of view about this particular change of variables, you may want to have a look at this post : math.stackexchange.com/questions/2133856/… In the end it all comes back to transform point by point the domain of integration and see what it looks like. In this case it is simple because the original domain of integration is an isosceles triangle and it transforms into another isosceles triangle. – zwim Jun 7 '17 at 13:43

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 $$\int_{x=0}^1\left( \int_{y=0}^{1-y} \cos\left(\frac{x-y}{x+y}\right)dy\right)dx=\frac{1}{2}\int_{u=0}^1\left(\int_{v=-u}^u\cos(v/u)dv\right) du.$$ Can you take it from here?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}