I was trying to find the normalization constant of the following distribution: p ( x ) = ( 1 − ( 1 − q ) x 2 ) 1 1 − q where 1 &lt; q &lt; 3

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I was trying to find the normalization constant of the following distribution:  p  (  x  )  =  (  1  −  (  1  −  q  )      x    2        )                  1                  1          −          q                    where   1  &amp;lt;  q  &amp;lt;  3

furniranizq · Accepted Answer

Effectively, the antiderivative express in terms of the gaussian hypergeometric function  I  =  ∫            [        1  −  (  1  −  q  )      x    2    )                    ]                            1                  1          −          q                        d  x  =  x          2        F    1        (          1      2        ,          1              q        −        1              ;          3      2        ;    (    1    −    q    )          x      2        )  and the definite integral  J  =      ∫          −      ∞              +      ∞                  [        1  −  (  1  −  q  )      x    2    )                    ]                            1                  1          −          q                        d  x  =            π              q        −        1                            Γ              (                              3            −            q                                2            (            q            −            1            )                          )                    Γ              (                  1                      q            −            1                          )            provided that  2  ℜ      (                  q        −        2                    q        −        1              )    &amp;lt;  1  ∧  (  ℜ  (  q  )  ≥  1  ∨  q  ∉      R    )which is your case with   1  &amp;lt;  q  &amp;lt;  3For the second integral, I do not see any possible solution except a series expansion around c=0For simplicity, let   a  =  q  −  1 and b=      1          1      −      q       to make  K  =      ∫          −      ∞              +      ∞                  (      1      +      a              x        2            )        b              (      1      +      a      (      x      +      c              )        2            )        b      d  xusing            (      1      +      a      (      x      +      c              )        2            )              b        =      ∑          n      =      0        ∞        Q    n          c    n    K  =      ∑          n      =      0        ∞        c    n          K    n      where        K    n    =      ∫          −      ∞              +      ∞                  (      1      +      a              x        2            )        b          Q    n      d  xWe have       K          2      n      +      1        =  0 and      K          2      n        =  (  −  1      )                  n        2                  a                            n          −          1                2                  π                 Γ              (                              n            −            1                    2                −        2        b        )                    Γ              (                  n          2                +        1        )            Γ      (      n      −      2      b      )              ∏          k      =      0                      n        2            −      1        (  b  −  k      )    2  This means that K is again an hypergoemetric function.

I was trying to find the normalization constant of the following distribution: p ( x ) =

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