There is a way to solve this problem?
Let be $[a,b]$ an interval where $a$ is finite but $b$ can also be infinity. Find a function or a distribution $h(u,s)$ for $u,s \in \mathbb{R}$ such that for any $n\in \mathbb{N}$ and any real function $F(s)$
$$ \int_{a}^{b}du \frac{h(u,s)}{(u-m^2)^{n+1}} = 1 = \int_a^b ds \frac{h(u,s)}{(s-m^2)^{n+1}} \tag1 $$ and $$ \int_{a}^{b}du\int_a^b ds \frac{h(u,s)F(s)}{(u-m^2)^{n+1}(s-m^2)^{n+1}} = \int_{a}^{b} ds \frac{F(s)}{(s-m^2)^{n+1}} \tag2. $$ (2) tells $h$ has to be symmetric under the exchange of $u$ and $s$, i.e. $h(u,s) = h(s,u)$. I tried a lot of times, but I don't menage to include the constraint of symmetry.
EDIT (after comments) Sorry, I edited the question correcting the mistake. Here, I add more informations. Notice that $n\in \mathbb{N}$.
- $m^2$ can take any value. For convenience, I will use $m^2\sim 0$ but it is not necessary.
- $m^2< a$ always
- I would like to find a $h$ independent of $n$.
- I prefer to handle with distributions, but also functions are allowed.