Solving ∫ 0 ∞ ( 1 + y 1 2 + y 2 2 + ⋯ + y n 2 ) ν ) d y 1 d y 2 ⋯ d y n

Question

Solving       ∫    0    ∞        (    1    +                            y                      1                                2                          +                  y                      2                                2                          +        ⋯        +                  y                      n                                2                          )            ν        )        d        y          1            d        y          2        ⋯      d        y          n

ordonansexa · Accepted Answer

As you have noticed, the integral can be transformed into      ∫    0    ∞        d    n    y      1          (      1      +              y        T            y              )        α              .Going to the spherical coordinates leads to      1          2      n        ∫  d      Ω    n        ∫    0    ∞    d  r            r              n        −        1                    (      1      +              r        2                    )        α              ,where the factor of   1      /        2    n   is due to integrating over this fraction of the whole space and   ∫  d      Ω          n       is the volume of       S          n      −      1      , which is equal to  ∫  d      Ω          n        =            2              π                  n                      /                    2                            Γ              (                  n          2                )              ,as can be shown for example by evaluating the integral   ∫      d    n    y      e          −              y        T            y       both in Cartesian and spherical coordinates and comparing the two results.To evaluate the simple integral over r, make the change of variables   x  =  (  1  +      r    2        )          −      1      , which leads to      ∫    0    ∞    d  r            r              n        −        1                    (      1      +              r        2                    )        α              =      1    2        ∫    0    1        x          α      −      n              /            2      −      1        (  1  −  x      )          n              /            2      −      1        d  x  =            Γ      (      α      −      n              /            2      )      Γ      (      n              /            2      )              2      Γ      (      α      )        ,where the expression for the Beta function was used in the last step. Putting the pieces together, you arrive at      ∫    0    ∞        d    n    y      1          (      1      +              y        T            y              )        α              =            (              π        4            )              n              /            2                  Γ      (      α      −      n              /            2      )              Γ      (      α      )        .

Solving int_0^infty (1+(y_1^2+y_2^2+...+y_n^2)/v dy_1dy_2... dy_n

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