Orthogonal Polynomials on (0,1) with kernel 1/x

I don’t know whether it’s a known result.

Let $\{P_n(x)\}$ be orthogonal polynomials on $(0,1)$ with measure $d\log x$ such that $$\langle P_m,P_n\rangle := \int_0^1 d\log x\, P_m(x)P_n(x)=\frac{\delta_{mn}}{2m},$$ with $P_1(x)=x$ , $P_n(0)=0$ and $P_n(1)=1$ .

We can directly construct $\{P_n\}$ from $x, x^2, \cdots$ by the Gram–Schmidt process, which gives us a recursion relation $$x\frac{d}{dx}(P_{n+1}-P_n)=(n+1)P_{n+1}+nP_n,$$ and it’s not hard to prove that $$P_n(x)=\sum_{k=1}^n(-1)^{n-k}{n\choose k}{n+k-1\choose n} x^k.$$ It’s also easy to prove the following identities: $$Q_n(x):=\int_0^x d\log t\, P_n(t)=\frac{(-1)^{n-1}}{n}(1-P_n(1-x)),$$ $$\langle P_n,x^k\rangle =\frac{1}{n+k}\prod_{i=1}^{k-1}\frac{n-i}{n+i}=\frac{(n-1)\cdots (n-k+1)}{(n+1)\cdots (n+k)},\quad \langle P_n,1\rangle =\frac{(-1)^{n+1}}{n},$$ $$(x\partial_x+n) P_n=2\sum_{k=1}^n k P_k,\quad n Q_n+P_n=2\sum_{k=1}^n (-1)^{n-k} P_k,$$ $$n\langle \log(x),P_n\rangle = -n\langle Q_n,1\rangle =(-1)^{n+1}\biggl(\frac{1}{n}-2\sum_{k=1}^n\frac{1}{k}\biggr),\quad \langle \log(1-x),P_n(x)\rangle = -\frac{1}{n^2},$$ $$n\langle x\log x,P_n(x)\rangle = \frac{(-1)^{n} }{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right),\quad \langle x\log x,P_1(x)\rangle=-\frac 14$$

Buwai Lee

交换图都不会画的魔法师