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I try to find the coefficients for Stieltjes polynomials with respect to Legendre polynomials. In other words, I wanna solve a problem for $n+1$ order polynomial $F_{n+1}(x)=a_{n+1}x^{n+1}+a_nx^n+...+a_1x+a_0$ : $$\int_{-1}^1 P_n(x)F_{n+1}(x)x^j dx=0 $$ where $P_n$ is the $n$th order Legendre polynomial and $j=0,1,...,n$ (note that this has a one degree of freedom, anyway we can solve it by specifying leading coefficients of $F_{n+1}$).

By substituting $j=0$, using the property of $P_n$ we have $$\int_{-1}^1 a_{n+1}P_n(x)x^{n+1}+a_nP_n(x)x^ndx=0 $$ and taking $j=0$ to $n$, these involves a calculation for $$\int_{-1}^1 P_n(x)x^kdx, \quad k=n,n+1,...,2n+1$$ Therefore the problem is reduced to calculate $$\int_{-1}^1 (\frac{d^n}{dx^n}(x^2-1)^n) x^kdx, \quad k=n,n+1,...,2n+1$$ or equivalently, after ignoring constants, $$\int_{-1}^1 (x^2-1)^n x^jdx, \quad j=0,1,...,n+1$$ I tried to substitue $x=cos(\theta)$ but this yields another complicated problem $$\int_0^\pi sin^{2n+1}(x)cos^j(x)dx, \quad j=0,...,n+1$$ Is there any convenient way to solve at least one of the last two equations? Thanks for any helps

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Note $[-\frac{1}{j+1}\cos^{j+1}(x)]'=\sin(x)\cos^j(x)$. Therefore, by using integration by parts, we have $$I=\int_0^\pi \sin^{2n}(x)\sin(x)\cos^j(x)dx=-\int_0^\pi (2n)\sin^{2n-1}(x)\cos(x)(-\frac{1}{j+1})\cos^{j+1}(x)dx=\frac{2n}{j+1}\int_0^\pi\sin^{2n-1}(x)\cos^j(x)(1-\sin^2(x))dx $$ Therefore, we have $$(1+\frac{2n}{j+1})I=\frac{2n}{j+1}\int_0^\pi \sin^{2n-1}(x)\cos^j(x)dx$$ $$I=\frac{2n}{2n+j+1}\int_0^\pi \sin^{2n-1}(x)\cos^j(x)dx$$ Hence, after do this $n$-times, we have $$I=\frac{2n}{2n+j+1}\frac{2n-2}{2n-2+j+1}...\frac{2}{2+j+1}\int_0^\pi \sin(x)\cos^j(x)dx$$ and the last integral is a derivative of $-\frac{1}{j+1}\cos^{j+1}(x)$.

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