This is a problem of ordinary calculus. Given \begin{align} f(x)&= \left\{ \begin{array}{ccc} k(0),&\quad 0\leq x<b\\ k(x-b),& \quad b\leq x<b+a \end{array} \right.,\\ g(x)&= \left\{ \begin{array}{cc} k(x),&\quad 0\leq x<a\\ k(a),& \quad a\leq x<b+a \end{array} \right.. \end{align} where $k(x)$ is a continuous function of $x\in[0,a]$ and $a,b>0$ $(a\neq b)$. If the values of $b$, $k(0)$, $k(a)$ $(k(a)\neq k(0)$ are fixed, do the following integrals separately have fixed values for arbitrary $a$ and $k(x)$ ($x\in(0,a)$)? How to prove it?
\begin{align} I_{1}&=\int^{a+b}_{0}dx\int^{x}_{0}dy [f(x)-g(x)]\cos(x-y)[f(y)-g(y)],\\ I_{2}&=\int^{a+b}_{0}dx\int^{x}_{0}dy [f(x)-g(x)] \sin(x-y)[f(y)+g(y)]. \end{align}
This is true if $k(x)$ is a step function with the jump at any $x\in[0,a)$. For general case, I have tried to do translation for $g(x)$, however, still can not prove it.
