Energy integral using Chebyshev coefficients I am trying to evaluate the following integral
\begin{equation}
E = \int_{-1}^1 f(x)g(x) \ dx
\end{equation}
I have the values for Chebyshev coefficients of the function $f(x) = \sum_{m=0}^{N-1} \hat{f}(m) \ T_m (x)$ and $g(x) = \sum_{m=0}^{N-1} \hat{g}(m) \ T_m (x)$ where $T_m(x)$ are Chebyshev polynomials of the first kind.
The discrete and continuous orthogonality relations on the Chebyshev polynomials are  given by the following where in the discrete orthogonality relation $\{ x_i \}$ are the roots of the Chebyshev polynomial of the highest order i.e, $T_{N_r-1}(x)$:
\begin{equation}
 \sum_i T_n(x_i) T_m(x_i) =
 \begin{cases}
 0 & \quad n \neq m \\
 N_r/2 & \quad n=m \neq 0 \\
 N_r & \quad n=m=0 \\
\end{cases}
\label{eq:disc_orth_cheb}
\end{equation}
\begin{equation}
 \int_{-1}^1 T_n(x) T_m(x) \frac{dx}{\sqrt{1-x^2}} =
 \begin{cases}
 0 & \quad n \neq m \\
 \pi/2 & \quad n=m \neq 0 \\
 \pi & \quad n=m=0 \\
\end{cases}
\label{eq:cont_orth_cheb}
\end{equation}
Is there any way I can compute the integral $E$ in terms of $\hat{f}(m)$ and $\hat{f}(n)$?
The following attempt is incomplete since I am not sure the integral to sum conversion is complete. There is a missing factor but I cannot figure out what it can be..
\begin{align*}
 E =& \int_{-1}^1  \ f(x)^2 \ dx \\
 =& \sum_{m} \sum_{n} \sum_{i (?)} \hat{f}(m) \hat{f}(n) T_m(x_i) T_n(x_i) \\
 =& \sum_{m} \sum_{n} \hat{f}(m) \hat{f}(n)  \sum_{i (?)}T_m(x_i) T_n(x_i)  \end{align*}
Applying discrete orthogonality relation, we can get the following
\begin{align*}
E=& \frac{N_r}{2} \sum_{m=0}^{N_r -1}g(m) \hat{f}(m)^2 
\end{align*}
However, this relies on resolving the $(?)$ that arises when we write the definite integral in $x$ in terms of the sum over the value of the integrand at the discrete collocation points which are the Chebyshev roots $x_i$.
 A: Let's plug the expansion
$$
f(x) = \sum_{m=0}^{n-1} {\hat f}_m T_m(x)\\
g(x) = \sum_{m=0}^{n-1} {\hat g}_m T_m(x)\\
$$
into the bilinear form (with some fixed $\omega(x)$ weight function)
$$
J(f, g) = \int_{-1}^1 f(x) g(x) \omega(x) dx = 
\int_{-1}^1 
\left(
\sum_{m=0}^{n-1} {\hat f}_m T_m(x)
\right) \cdot \left(
\sum_{k=0}^{n-1} {\hat g}_k T_k(x)
\right)
\omega(x)
dx = \\
= \sum_{m,k=0}^{n-1} {\hat f}_m {\hat g}_k \int_{-1}^1 T_m(x) T_k(x) \omega(x) dx = 
\sum_{m,k=0}^{n-1} a_{mk} {\hat f}_m {\hat g}_k.
$$
Here
$$
a_{mk} = \int_{-1}^1 T_m(x) T_k(x) \omega(x) dx, \quad m, k = 0,\dots,n-1
$$
is a matrix which elements can be computed if $\omega(x)$ is known.
If $\omega(x) = \frac{1}{\sqrt{1-x^2}}$ then
$$
a_{mk} = \begin{cases}
0, & m\neq k\\
\frac{\pi}{2}, &m = k > 0\\
\pi, &m = k = 0
\end{cases}.
$$
So
$$
\int_{-1}^1 \frac{f(x)g(x)}{\sqrt{1-x^2}} dx = \pi f_0 g_0 + \frac{\pi}{2}
\sum_{m=1}^n f_m g_m.
$$
If $\omega(x) = 1$ (as in your original question) then
$$
a_{mk} = \int_{-1}^1 T_m(x) T_k(x) dx.
$$
This matrix does not have a simple expression like above. But a closed form still exists:
$$
T_m(x) T_k(x) = \cos m \arccos x \cos k \arccos x = \\ =
\frac{1}{2} \left(
\cos (m-k) \arccos x + \cos (m+k) \arccos x
\right)
= \frac{T_{|m-k|}(x) + T_{m+k}(x)}{2}
$$
The $T_m(x)$ can be easily integrated (see the last property here):
$$
\int_{-1}^1 T_m(x) dx = \begin{cases}
\frac{2}{1 - m^2}, &m \text{ is even}\\
0, &m \text{ is odd}
\end{cases}
$$
So
$$
a_{mk} = \begin{cases}
\frac{1}{1 - (m-k)^2} + \frac{1}{1 - (m+k)^2}, & m - k \text{ is even}\\
0, & m - k \text{ is odd}
\end{cases}
$$
Collecting everything we get
$$
\int_{-1}^1 f(x) g(x) dx = \sum_{\substack{k,m=0\\k + m \text{ is even}}}^{n-1}
\left[
\frac{1}{1 - (m-k)^2} + \frac{1}{1 - (m+k)^2}
\right]
{\hat f}_m {\hat g}_k.
$$
The bad news is that evaluating $\int_{-1}^1 f(x) g(x) dx$ requires $O(n^2)$ operations if only ${\hat f}_m, {\hat g}_m$ are known, while evaluating $\int_{-1}^1 \frac{f(x) g(x)}{\sqrt{1-x^2}} dx$ requires only $O(n)$.
