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It can be proved by induction that the recurrence relation $$K_n(u,v)=\int_0^1K_1(u,u_1)K_{n-1}(u_1,v)du_1,$$ with $K_1(u,v)=\theta(1-u-v)$, where $\theta(x)$ is the Heaviside step function with $\theta(0)=1/2$, has the solution (see http://arxiv.org/abs/1207.2055 ) $$K_{2n}(u,v)=(-1)^n\frac{2^{2n-2}}{(2n-1)!}\left ( \left [ E_{2n-1}\left ( \frac{u+v}{2}\right )+E_{2n-1}\left (\frac{u-v}{2}\right )\right ]\theta(u-v)+\right . $$ $$\left . \left [ E_{2n-1}\left (\frac{u+v}{2}\right )+ E_{2n-1}\left (\frac{v-u}{2}\right )\right ]\theta(v-u) \right )$$ and $$K_{2n+1}(u,v)=(-1)^n\frac{2^{2n-1}}{(2n)!}\left ( \left [ E_{2n}\left ( \frac{1-u+v}{2}\right )+E_{2n}\left (\frac{1-u-v}{2}\right )\right ]\theta(1-u-v)+\right . $$ $$\left . \left [ E_{2n}\left (\frac{1-u+v}{2}\right )- E_{2n}\left (\frac{u+v-1}{2}\right )\right ]\theta(u+v-1)\right ) .$$ In these formulas $E_n(x)$ are the Euler polynomials.

These expressions have been simply guessed. How can these results be obtained by some systematic method?

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So what is the question? –  Per Alexandersson Apr 5 '13 at 14:47
    
The question is how this solution can be obtained by some regular method without guessing it (which was not a simple task). –  Zurab Silagadze Apr 5 '13 at 16:36

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