<s>[INST] Finite Riemann-like sum and integration For all $x<0$, $y>0$, let's define the following function on $[-1,1]$ : $$ f_{x,y}(t) = \frac{-2xy}{((y-x) - t*(x+y))^2} $$
The denominator is never zero, so it is well defined. It is specially designed so that (you may verify if you want) : $$ \int_{t=-1}^1 f_{x,y}(t) dt = 1$$
My question is : Does there exist a finite set of abscissae $\{t_0,\dots,t_n\}\in[-1,1]^n$ and positive weights $\{w_0,\dots,w_n\} \in \mathbb{R}^n$ (independent of $x$ and $y$) such that for all values $x<0$ and $y>0$ we have : $$ 1 = \int_{t=-1}^1 f_{x,y}(t) dt  = \sum_{i=1}^n w_i f_{x,y}(t_i)$$
Any tip, partial of full solution is welcome.
 [/INST] WLOG, we assume $x+y\ne 0$.
\begin{align}
\int_{-1}^1\frac{-2xy}{((y-x)-t(x+y))^2}\,dt
&=
\frac{-2xy}{(x+y)^2}\int_{-1}^1\frac{dt}{(t+\frac{x-y}{x+y})^2}
\\ &=
\frac{-2xy}{(x+y)^2}\int_{-1+\frac{x-y}{x+y}}^{1+\frac{x-y}{x+y}}\frac{dt}{t^2}
\end{align}
Put $x_i=-1+\frac{x-y}{x+y}+\frac{2i}{n}$ and $\xi_i=\sqrt{x_{i=1}x_i}$ . Note that
\begin{align}
\int_{-1+\frac{x-y}{x+y}}^{1+\frac{x-y}{x+y}}\frac{dt}{t^2}
&=
\lim_{n\to \infty} \sum_{i=1}^n \frac{x_i-x_{i-1}}{x_{i-1}x_i}
\\ &=
\lim_{n\to \infty} \sum_{i=1}^n \left( \frac{1}{x_{i-1}}-\frac{1}{x_{i}} \right)
\\ &=
\lim_{n\to \infty} \left( \frac{1}{x_{0}}-\frac{1}{x_{n}} \right)
\\ &=
\frac{1}{-1+\frac{x-y}{x+y}}-\frac{1}{1+\frac{x-y}{x+y}}
\end{align}
Thus
\begin{align}
\int_{-1}^1\frac{-2xy}{((y-x)-t(x+y))^2}\,dt
&=
\frac{-2xy}{(x+y)^2}\int_{-1+\frac{x-y}{x+y}}^{1+\frac{x-y}{x+y}}\frac{dt}{t^2}
\\ &=
\frac{-2xy}{(x+y)^2}\frac{2}{(\frac{x-y}{x+y})^2-1}
\\ &=
1
\end{align}
</s>