Why is the sum of chebyshev gaussian quadraure weights not $2$? When I sum the weights of the chebyshev gaussian quadrature over the interval $[-1,1]$. I get about $1.57$. I don't understand why it is not equal to the domain size of the integral (i.e. $2$). My question is why?
 A: Chebyshev-Gauss quadrature, also called Chebyshev quadrature, is a Gaussian quadrature over the interval $[-1,1]$ with weighting function $W(x)=\frac{1}{\sqrt{1-x^{2}}}$. The abscissas for quadrature order $n$ are given by the roots of the Chebyshev polynomial of the first kind $T_n(x)$, which occur symmetrically about $0$. The weights are
$$w_i = -\frac{A_{n+1}\gamma_n}{A_nT'_n(x_i)T_{n+1}(x_i)}...(1) $$
    $$ = \frac{A_n}{A_{n-1}}\frac{\gamma_(n-1)}{T_{n-1}(x_{i})T'_n(x_i)}...(2)$$, where $A_n$ is the coefficient of $x^n$ in $T_n(x)$,
$$ \gamma_n = A_n\pi(x)...(3)$$
and $\pi(x)$ the order-$n$ Lagrange interpolating polynomial for $T_n(x)$.
For Chebyshev polynomials of the first kind,
$$ A_n=2^{n-1}...(4)$$
so
$$\frac{A_{n+1}}{A_n}=2...(5)$$
Additionally,
$$\gamma_n=\frac{1}{2\pi}...(6)$$
so
$$ w_i=-\frac{\pi}{T_{n+1}(x_i)T'_n(x_i)}...(7)$$
Since
$$ T_n(x)= \cos(ncos^{-1}x)...(8)$$
the abscissas are given explicitly by
$$ x_i=\cos[\frac{(2i-1)\pi}{2n}]...(9)$$
Since
$$ T'_n(x_i) = \frac{(-1)^{i+1}n}{\sin\alpha_i}...(10)$$ and
$$ T_{n+1}(x_i) = (-1)^{i}\sin\alpha_i...(11)$$
where
$$ \alpha_i=\frac{(2i-1)\pi}{2n}...(12)$$
all the weights are
$$w_i= \frac{\pi}{n}.$$ Hope it helps.
