Can any one help me to show that $ u(x,y) = \log |(x+y)/(x-y)| $ is locally integrable on $R^{2}$ ? I guess yes because it only can problem near $y= x$. Further how to find its distributional derivative $u_{xy}$.
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If we make a linear change of variable $z=x+y$, $x-y$, then our function $u$ becomes $\log|z|-\log|w|$. Note that $\log|z|$ is locally integrable on $\mathbb{R}$, thus it's also locally integrable on $\mathbb{R}^2$. Similarly with $\log|w|$. Since a linear change of variable does not change local integrability, $u$ is locally integrable in the original variables $x,y$.
Now we begin to compute $u_{xy}$. Note that for $\varphi\in C_c^\infty(\mathbb{R}^2)$, $\varphi=\varphi(z,w)$, we have the following identity: $$\frac{\partial^2}{\partial x\partial y}\varphi(x+y,x-y)= \varphi_{zz}(x+y,x-y)-\varphi_{ww}(x+y,x-y).$$ This identity extends to distributions by duality. Now write $$u(x,y)=f(x+y,x-y)-g(x+y,x-y)=\log|x+y|-\log|x-y|$$ for $$f(z,w)=\log|z|$$ and $$g(z,w)=\log|w|,$$ then we have $$u_{xy}(x+y,x-y)=f_{zz}(x+y,x-y)+g_{ww}(x+y,x-y).$$ Let's now compute $f_{zz}$. Note that $\log|z|$ defines a distribution on $\mathbb{R}$, and let's denote this distribution by $a$. Then $f=a\otimes \text{Id}$ where $\text{Id}$ is the identity distribution on the $w$ variable. Thus $f_{zz}=a''\otimes\text{Id}$. Similarly, $g_{ww}=\text{Id}\otimes a''$, and thus $u_{xy}=a''\otimes\text{Id}+\text{Id}\otimes a''$.
Now for $a(z)=\log|z|$, $\varphi\in C_c^\infty(\mathbb{R})$, write $$(a,\varphi')=\int \log|z| \varphi(z)\ dz =\lim_{\varepsilon\to 0+}\int_{|z|>\varepsilon}\log|z|\varphi'(z)\ dz$$ and perform integration by part to get $$(a',\varphi)=\lim_{\epsilon\to 0+}\int_{|z|>\epsilon}\frac{\varphi(z)}{z}\ dz=\int_{|z|>1}\frac{\varphi(z)}{z}\ dz+\int_{|z|\leq 1}\frac{\varphi(z)-\varphi(0)}{z}\ dz.$$ Then $$(a'',\varphi) =-(a',\varphi')=-\int_{|z|>1}\frac{\varphi'(z)}{z}\ dz-\lim_{\varepsilon\to 0+}\int_{\varepsilon<|z|\leq 1}\frac{\varphi'(z)-\varphi'(0)}{z}\ dz.$$ By integration by parts again, we get $$(a'',\varphi)=-\int_{|z|>1}\frac{\varphi(z)}{z^2}\ dz- \lim_{\varepsilon\to 0+}\int_{\varepsilon<|z|\leq 1} \frac{\varphi(z)-\varphi'(0)z-\varphi(0)}{z^2}\ dz+2\varphi'(0).$$
So in short, $u_{xy}=a''\otimes\text{Id}+\text{Id}\otimes a''$. Explicitly, for $\varphi\in C_c^\infty(\mathbb{R}^2)$, $$(u_{xy},\varphi)=(a''\otimes\text{Id},\varphi)+(\text{Id}\otimes a'',\varphi)$$ where $$(a''\otimes\text{Id},\varphi) =-\int_{|x+y|>1}\frac{\varphi(x,y)}{(x+y)^2}\ dxdy- \lim_{\varepsilon\to 0+}\int_{\varepsilon<|x+y|\leq 1} \frac{\varphi(x,y)-\frac{1}{2}(\varphi_x(\frac{x-y}{2},\frac{y-x}{2})+\varphi_y(\frac{x-y}{2},\frac{y-x}{2}))(x+y)-\varphi(\frac{x-y}{2},\frac{y-x}{2})}{(x+y)^2}\ dxdy+\varphi_x(\frac{x-y}{2},\frac{y-x}{2})+\varphi_y(\frac{x-y}{2},\frac{y-x}{2})$$ and $$(\text{Id}\otimes a'',\varphi) =-\int_{|x-y|>1}\frac{\varphi(x,y)}{(x-y)^2}\ dxdy- \lim_{\varepsilon\to 0+}\int_{\varepsilon<|x-y|\leq 1} \frac{\varphi(x,y)-\frac{1}{2}(\varphi_x(\frac{x+y}{2},\frac{x+y}{2})-\varphi_y(\frac{x+y}{2},\frac{x+y}{2}))(x-y)-\varphi(\frac{x+y}{2},\frac{x+y}{2})}{(x-y)^2}\ dxdy+\varphi_x(\frac{x+y}{2},\frac{x+y}{2})-\varphi_y(\frac{x+y}{2},\frac{x+y}{2}).$$
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