How to compute ∫ a b ( b − x ) n − 1 2 ( x − a ) − 1 / 2 d x

Question

How to compute       ∫    a    b    (  b  −  x      )                            n          −          1                2              (  x  −  a      )          −      1              /            2        d  x

Demarion Thornton · Accepted Answer

I  =      ∫    a    b                      b        −        x                            n                    (      b      −      x      )      (      x      −      a      )          d  x                    I                            =                  ∫          0                      π                          /                        2                                                2            (            b            −            a            )            cos            ⁡            t            sin            ⁡            t            ⋅                                          (                b                −                a                )                                  cos                                      2                                                  ⁡                t                                                            n                                                          (            b            −            a                          )              2                                      cos              2                        ⁡            t                          sin              2                        ⁡            t                                  d        t                                          =        2                              b            −            a                                            n                                    ∫          0                      π                          /                        2                                    cos                      n                          ⁡        t                d        t                                          =                  {                                                    π                                                      b                    −                    a                                                                            n                                                                                        1                    ⋅                    3                    ⋯                    (                    n                    −                    1                    )                                                        2                    ⋅                    4                    ⋯                    n                                                                  (                n                                even                )                                                                    2                                                      b                    −                    a                                                                            n                                                                                        2                    ⋅                    4                    ⋯                    (                    n                    −                    1                    )                                                        1                    ⋅                    3                    ⋯                    n                                                                  (                n                                odd                )                                                                  The latter integral is well known, and can easily be evaluated by parts:                              I          n                =                  ∫          0                      π                          /                        2                                    cos          n                ⁡        t                d        t        ⇒                  I          n                                    =        (        n        −        1        )                  ∫          0                      π                          /                        2                                    sin          2                ⁡        t                  cos                      n            −            2                          ⁡        t                d        t                                          =        (        n        −        1        )                  I                      n            −            2                          −        (        n        −        1        )                  I                      n                                                                    ⇒                  I          n                =                              n            −            1                    n                          I                      n            −            2                                                                            ⇒                  {                                                                      I                  n                                =                                                      1                    ⋅                    3                    ⋯                    (                    n                    −                    1                    )                                                        2                    ⋅                    4                    ⋯                    n                                                                    I                  0                                                (                n                                even                )                                                                                      I                  n                                =                                                      2                    ⋅                    4                    ⋯                    (                    n                    −                    1                    )                                                        1                    ⋅                    3                    ⋯                    n                                                                    I                  1                                                (                n                                odd                )                                                                                                                                              (                                    I          0                =                              π            2                                                            I          1                =        1                              )

How to compute int_a^b (b-x)^(n-1/2)(x-a)^(-1/2)dx?

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