<s>[INST] orthogonal polynomials in (-1,1) with a modified weight function I would like to find the orthogonal polynomial system   $ \{ P_n(x), n \in \textbf{Z} \} $ corresponding to the weight function $ w(x) = \frac { 1} {\sqrt{1-x^2} (1+\sqrt{1-x^2} ) } $  defined on the interval $(-1,1)$. Noticing that this $w(x)$ is a modified Chebyshev weight function, I am just wondering whether the solution to this problem studied previously. I am not sure about it.  Your responses are very much appreciated. 
 [/INST] The integral $ \int_{x=-1}^{+1} \frac {dx} {\sqrt{1-x^2} (1 + \sqrt{1-x^2}) } = \int_{x=-1}^{0} \frac {dx} {\sqrt{1-x^2} (1 + \sqrt{1-x^2} )} + \int_{x=0}^{1} \frac {dx} {\sqrt{1-x^2} (1 + \sqrt{1-x^2} )} $.
Using  substitutions $ s \rightarrow -\sqrt{1-x^2}$ and $t \rightarrow\sqrt{1-x^2} $ respectively in to the first and second integrals on the right-hand side gives
$ \int_{x=-1}^{+1} \frac {dx} {\sqrt{1-x^2} (1 + \sqrt{1-x^2}) } = 
 - \int_{s=-1}^{0} \frac{ds}{(1+s)^{ \frac{1}{2}} (1-s)^{\frac{3}{2}}} +
  \int_{t=0}^{1} \frac{dt}{(1+t)^{ \frac{3}{2}} (1-t)^{\frac{1}{2}}} $ .
Thus, the resulting polynomial space appears to be
$L_2[(-1,1),w(x)] =L_2[(-1,0),w_1] \cup L_2[(0,1),w_2] $.
Where 
$ w(x) = \frac{1} { \sqrt{1-x^2} (1+\sqrt{1-x^2} ) }$  ,
$w_1(x) = \frac{-1} { (1+x)^{1/2} (1-x)^{3/2 } } $ and
$w_2(x) = \frac{1} { (1+x)^{3/2} (1-x)^{1/2 } } $.
So two different beta distributions for the half-intervals of $(-1,1)$, leading to two different orthogonal polynomial sequences there.
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