Finite Riemann-like sum and integration

Question

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.

Answer 1

Substituting $x=-\alpha y$, we have $f_{x,y}(t) = \frac{2\alpha}{(1+\alpha -(1-\alpha)t)^2}$. The question can now be stated for all $\alpha >0$. This is equivalent to ask for the equality of two rational functions with indeterminate $\alpha$. Its partial fraction decomposition wrt $\alpha$ is : $$ f_{x,y}(t) = \frac{\frac{2}{(1+t)^2}}{\alpha + \frac{1-t}{1+t}} - \frac{2\frac{1-t}{(1+t)^3}}{(\alpha + \frac{1-t}{1+t})^2}$$ We ask if there is a finite set of abscissae and weights such that $$1 = \sum_{i=1}^n w_i f_{x,y}(t_i)$$ But this expression would itself be its partial fraction decomposition wrt $\alpha$. Uniqueness of the decomposition thus implies that such abscissea and weights cannot exist.

Answer 2

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}