Double integral $\int_{[0,1]^2} \frac{f(x)-f(y)}{x-y} dx\,dy$ for regular function $f$

Question

Let $f$ be a sufficiently regular function on the unit square. (For example, say $f\in C^1$.) Then, is there a way to simplify the integral $$\int_{[0,1]^2} \frac{f(x)-f(y)}{x-y} dx\,dy ?$$ If $f\in C^1$, then the integrand is bounded and therefore the integral is well-defined.

Answer 1

Denote the quantity in question by $Q$. Due to symmetry it is sufficient to integrate over $\>T\!: \ 0\leq y<x\leq1$. Furthermore we write $$f(x)-f(y)=\int_y^xf'(t)\>dt\ .$$ We then have $$Q=2\int_T \left( {1\over x-y}\int_y^x f'(t)\>dt\right)\> {\rm d}(x,y)=\int_0^1 \left(\int_0^t \int_t^1{2\over x-y}\>dx\>dy\right) f'(t)\>dt\ .$$ We now have to compute the weight function $$\eqalign{h(t) &:= 2\int_0^t \int_t^1 {1\over x-y}\>dx\>dy\cr &=2\int_0^t \log(x-y)\biggr|_{x=t}^{x=1}\>dy=2\int_0^t \bigl(\log(1-y)-\log(t-y)\bigr)\>dy\ ,\cr} $$ so that one obtains $$h(t)=2 \left(t\log{1\over t}+(1-t)\log{1\over 1-t}\right)\ .$$ The plot of $h(t)$ is a symmetric bump over the interval $[0,1]$. The final answer therefore is $$Q=\int_0^1 h(t)\>f'(t)\>dt\ .$$