I have tried to solve the integral $${\int_0^1 U_n (x) U_m (x)x dx },$$
where ${U_n (x) }$ denotes Chebyshev polynomial of the second kind. Solving the integration and checking the result, I noticed that it does not match the ones I did analytically for $n=0$, $m=1$; and some other values, so, there must be something wrong. I need to use $x$ as weighting function, and not the other standard ones.
I would appreciate any help.
Mohd
Here's the integration:
Formulas were taken from: https://en.wikipedia.org/wiki/Chebyshev_polynomials
We need to solve this integration,
$$\int{U_n\left(x\right)U_m\left(x\right)x\ dx}$$ The term $U_n\left(x\right)U_m\left(x\right)$ can be written as:
$$U_n\left(x\right)U_m\left(x\right)=\sum^{m+n}_{p=m-n,\ \ \ step\ 2}{U_p(x)}$$
$$\int{U_n\left(x\right)U_m\left(x\right)x\ dx}=\ \sum^{m+n}_{p=m-n,\ step\ 2}{\int U_p\left(x\right)\ x\ dx}$$
Hence, the integration problem is reduced to solving, $$\int U_p\left(x\right)\ x\ dx$$
Let ${u=x,\ and\ dv=U}_p\left(x\right)dx,\ then,\ du=dx,\ and\ $
$$v=\int{U_p\left(x\right)\ dx=}{{T_{p+1}\left(x\right)}\over {p+1}}+C$$
Where C is constant, then
$$\int U_p\left(x\right)xdx=uv-vdu={{xT_{p+1}\left(x\right)}\over {p+1}}-\int {{T_{p+1}\left(x\right)}\over {p+1}}dx+\int Cdx$$ and when $p>1$, we have
$$\int T_p\left(x\right)dx={{1}\over {2}}\left({{T_{p+1}(x)}\over {p+1}}-{{T_{p-1}(x)}\over {p-1}}\right)$$
and for $p+1$ we can write,
$${Y_p(x)} = \int T_{p+1}\left(x\right)dx={{1}\over {2}}\left({{T_{p+2}\left(x\right)}\over {p+2}}-{{T_p\left(x\right)}\over {p}}\right)$$
when $p=0$ , and since $T_1\left(x\right)=x$, gives $\int T_1\left(x\right)dx=\int xdx={{x^2}\over {2}}$. Thus we can write
$$\int U_p\left(x\right)xdx={{xT_{p+1}\left(x\right)}\over {p+1}}-{{Y_{p}\left(x\right)}\over {p+1}} +Cx$$ Where
$$Y_p\left(x\right)={{x^2}\over {2}},\ \ {for}\ p=0$$ $$Y_p\left(x\right)={{1}\over {2}}\left({{T_{p+2}\left(x\right)}\over {p+2}}-{{T_p\left(x\right)}\over {p}}\right),\ \ {for}\ p>0$$
We know that $T_p\left(1\right)=1,\ T_p\left(0\right)={{cos} {{p\pi }\over {2}}\ }$, thus,
$$Y_p(0)=0, for p=0$$
$$Y_p\left(0\right)={{1}\over {2}}\left({{{{cos} {{(p+1)\pi }\over {2}}\ }}\over {p+2}}-{{{{cos} {{p\pi }\over {2}}\ }}\over {p}}\right),\ {for}\ p>0$$
$$Y_p\left(1\right)={{1}\over {2}}\ ,{\ for}\ p=0\ $$ $$Y_p\left(1\right)={{1}\over {2}}\left({{1}\over {p+2}}-{{1}\over {p}}\right),\ \ for\ p>0$$
We know that $T_p(1)=1$, $T_p(0)={cos ({{p\pi }/2)}}$, thus,
$$\int^1_0{U_n\left(x\right)U_m\left(x\right)x\ dx}={\left[\sum^{m+n}_{p=m-n,\ \ \ step\ 2}{{{xT_{p+1}\left(x\right)}\over {p+1}}-Y_p(x)/(p+1)}\right]}^1_0$$
We need to handle the extreme case where $p=0$, hence, $p$ starts from $m-n+2$, when $m=n$
$$\int^1_0{U_n\left(x\right)U_m\left(x\right)x\ dx}={\left[\sum^{m+n}_{p=m-n,\ \ \ step\ 2}{{{xT_{p+1}\left(x\right)}\over {p+1}}-Y_p(x)/(p+1)}\right]}^1-{\left[\sum^{m+n}_{p=m-n,\ \ \ step\ 2}{{{xT_{p+1}\left(x\right)}\over {p+1}}-Y_p(x)/(p+1)}\right]}_0$$
$$={\left[\sum^{m+n}_{p=m-n+2{\delta }_{mn},\ \ \ step\ 2}{{{1\times T_{p+1}\left(1\right)}\over {p+1}}-Y_p\left(1\right)/(p+1)}\right]}^1-{\left[\sum^{m+n}_{p=m-n+2{\delta }_{mn},\ \ \ step\ 2}{{{0{\times T}_{p+1}\left(0\right)}\over {p+1}}-Y_p\left(0\right)/(p+1))}\right]}_0$$
$$+\left({{T_{0+1}\left(1\right)}\over {1}}-Y_0\left(1\right)-({{0{\times T}_{0+1}\left(0\right)}\over {0+1}}-Y_0(0)\right){\delta }_{mn}$$
$$={\left[\sum^{m+n}_{p=m-n+2{\delta }_{mn},\ \ \ step\ 2}{{{1}\over {p+1}}-{{1}\over {2}}\left({{1}\over {p+2}}-{{1}\over {p}}\right)}\right]}+{\left[\sum^{m+n}_{p=m-n+2{\delta }_{mn},\ \ \ step\ 2}{{{1}\over {2(p+1))}}\left({{{{cos} {{(p+1)\pi }\over {2}}\ }}\over {p+2}}-{{{{cos} {{p\pi }\over {2}}\ }}\over {p}}\right)}\right]}$$
$$+(1-{{1}\over {2}}+0){\delta }_{mn}$$
$$=\sum^{m+n}_{p=m-n+2{\delta }_{mn},\ \ \ step\ 2}{{{1}\over {p+1}}+\left({{1}\over {p(p+2)}}\right)}-\sum^{m+n}_{p=m-n+2{\delta }_{mn},\ \ \ step\ 2}{{{1}\over {2(p+1))}}\left({{{{cos} {{(p+1)\pi }\over {2}}\ }}\over {p+2}}-{{{{cos} {{p\pi }\over {2}}\ }}\over {p}}\right)}$$
$$+{{1}\over {2}}{\delta }_{mn}$$
Which can be finally written as:
$$\int^1_0{U_n\left(x\right)U_m\left(x\right)x\ dx}={{1}\over {2}}{\delta }_{mn}+\sum^{m+n}_{p=m-n+2{\delta }_{mn},\ \ \ step\ 2}{{{1}\over {p+1}}+\left({{1}\over {p(p+2)}}\right)}-{{1}\over {2(p+1)}}\left({{{{cos} {{(p+1)\pi }\over {2}}\ }}\over {p+2}}-{{{{cos} {{p\pi }\over {2}}\ }}\over {p}}\right)$$
Test: $$U_n\left(x\right)U_m\left(x\right),\ for\ n=0,\ m=1;\ U_0\left(x\right){=1,\ U}_1\left(x\right)=2x$$
$${{\left.\int^1_0{U_0\left(x\right)U_1\left(x\right)x\ dx}=2\int^1_0{x^2\ dx}={{2x^3}\over {3}}\right]}_0}^1=2/3$$
But, the formula of the integration gives 3/4, using the below matlab code.