Is there an easy way to find orthogonal polynomials w.r.t. $w(x)=-\ln(x)$ on $(0, 1)$?

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I've been evaluating them recurcively using $$\varphi_{n}=x^n - \sum_{k=0}^{n-1} \frac{\langle x^{k}, \varphi_{k} \rangle_{w}}{\langle \varphi_{k}, \varphi_{k} \rangle_{w}}\cdot\varphi_{k}$$

Where $\langle f,g \rangle_{w}=\int_{0}^{1} -\ln|{x}|f(x)g(x)\: dx$.

The computations get very complicated for $n\ge 3$.

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Try the substitution $x= e^{-t}$. Then the inner product $<x^j, x^k>$ of the monomial pairs $x^j, x^k$ in their transformed format is $$ \int_{t=0}^{t=\infty} e^{-kt} e^{-jt} (t e^{-t}) dt$$

That integral can be evaluated sneakily by making use of the identity $ -\frac{d}{da} e^{-at} = t e^{-at}$ and setting $a=1$ afterward.

In detail $$<x^j, x^k> = (-\frac {d}{da}) \int_0^{\infty} e^{-(k+j+a) t} \ dt = -\frac {d}{da}( \frac {1}{j+k+a})|_{a=1} $$

$$= \frac{1}{(j+k+1)^2}$$

Then you can use the usual Gram-Schmidt method to convert these monomials into an orthonormal basis.

P.S. Using Mathematica (by another method) I got this small table that you can use to check your work. Good luckenter image description here