For a complex number z, How to evaluate ∫ 0 ∞ d x x 2 + ( 1 − z 2 x 2 ) 2

Question

For a complex number z, How to evaluate      ∫    0    ∞              d      x                      x        2            +      (      1      −              z        2                    x        2                    )        2

boursecasa2je · Accepted Answer

By the quadratic formula, the denominator of the integrand has roots                              r          1                ,                  r          2                ,                  r          3                ,                  r          4                                    =        ±                                            2                              z                2                            −              1              ±                              1                −                4                                  z                  2                                                                    2                              z                4                                                                                      =                              ±            1                                2                          z              2                                                (          i          ±                      4                          z              2                        −            1                    )                    and since the integrand is an even function,                              ∫          0          ∞                                      d            x                                              x              2                        +            (            1            −                          z              2                                      x              2                                      )              2                                                          =                  1          2                          ∫                      −            ∞                    ∞                                      d            x                                              x              2                        +            (            1            −                          z              2                                      x              2                                      )              2                                                                        =                  1          2                          ∫                      −            ∞                    ∞                                      1                          /                                      z              4                                                          x              4                        +            (            1            −            2                          z              2                        )                          x              2                                      /                                      z              4                        +            1                          /                                      z              4                                      d        x                                          =                  1                      2                          z              4                                                ∫                      −            ∞                    ∞                f        (        x        )        d        x            where  f  (  x  )  =      1          (      x      −              r        1            )      (      x      −              r        2            )      (      x      −              r        3            )      (      x      −              r        4            )        .Since f(z) is analytic in the upper half complex plane (except for a finite number of poles), and since f(z) vanishes faster than   1      /        z    2   for       |    z      |    →  ∞, the residue theorem gives      1          2              z        4                  ∫          −      ∞        ∞    f  (  x  )  d  x  =            π      i              z      4        ∑      r    e    s            fwhere ∑resf is the sum of the residues in the upper-half plane. The trick then is to determine which of the four poles are on the upper half plane. We define the roots as                              r          1                                    =                  1                      2                          z              2                                                (          i          +                      4                          z              2                        −            1                    )                                              r          2                                    =        −                  r          1                                              r          3                                    =                  1                      2                          z              2                                                (          i          −                      4                          z              2                        −            1                    )                                              r          4                                    =        −                  r          3                    If z is purely real, then       r    1   and       r    3   are in the upper half plane. Thus                    ∑                  r          e          s                                    f                            =                  1                      (                          r              1                        −                          r              2                        )            (                          r              1                        −                          r              3                        )            (                          r              1                        −                          r              4                        )                          +                  1                      (                          r              3                        −                          r              1                        )            (                          r              3                        −                          r              2                        )            (                          r              3                        −                          r              4                        )                                                            =        −        i                              z            4                    2                    after much algebra. Therefore we have                              ∫          0          ∞                                      d            x                                              x              2                        +            (            1            −                          z              2                                      x              2                                      )              2                                                          =                              π            i                                z            4                                                −            i                          z              4                                2                                                  =                  π          2                    which is independent of z (if z is purely real)! In the general case, however, compute the residues of the poles that are in the upper half plane, to be determined by the value of z.

For a complex number z. How to evaluate int_0^infty dx/(x^2+(1-z^2x^2)^2)

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