Evaluate ∫ 0 ∞ x 2 ln ⁡ x x 4 + x 2 + 1 d x by the residue theorem

Question

Evaluate       ∫    0    ∞                      x        2            ln      ⁡      x                      x        4            +              x        2            +      1        d  x by the residue theorem

oedfeuonbk203 · Accepted Answer

You can use Feynman&#039;s method to evaluate. Let  I  (  a  )  =      ∫    0    ∞              x      a                      x        4            +              x        2            +      1        d  x  .Then                    I        (        a        )                            =                              ∫          0          1                                                    x              a                        +                          x                              2                −                a                                                                        x              4                        +                          x              2                        +            1                          d        x                                          =                              ∫          0          1                                      (            1            −                          x              2                        )            (                          x              a                        +                          x                              2                −                a                                      )                                1            −                          x              6                                      d        x                                          =                              ∫          0          1                          ∑                      n            =            0                    ∞                (        1        −                  x          2                )        (                  x          a                +                  x                      2            −            a                          )                  x                      6            n                          d        x                                          =                              ∑                      n            =            0                    ∞                          ∫          0          1                (        1        −                  x          2                )        (                  x          a                +                  x                      2            −            a                          )                  x                      6            n                          d        x                                          =                              ∑                      n            =            0                    ∞                          ∫          0          1                (                  x                      6            n            +            a                          +                  x                      6            n            +            2            −            a                          −                  x                      6            n            +            2            +            a                          −                  x                      6            n            +            4            −            a                          )        d        x                                          =                              ∑                      n            =            0                    ∞                                      (                                    1                      6            n            +            a            +            1                          +                  1                      6            n            +            3            −            a                          −                  1                      6            n            +            3            +            a                          −                  1                      6            n            +            5            −            a                                                )                              and hence                              I          ′                (        2        )                            =                              ∑                      n            =            0                    ∞                                      (                          −                  2                      (            6            n            +            3                          )              2                                      +                  1                      (            6            n            +            1                          )              2                                      +                  1                      (            6            n            +            5                          )              2                                                            )                                                            =                    −                  2          9                          ∑                      n            =            0                    ∞                          1                      (            2            n            +            1                          )              2                                      +                  ∑                      n            =            0                    ∞                                      (                                    1                      (            6            n            +            1                          )              2                                      +                  1                      (            6            n            +            5                          )              2                                                            )                                                            =                    −                  2          9                          ∑                      n            =            0                    ∞                          1                      (            2            n            +            1                          )              2                                      +                  ∑                      n            =            −            ∞                    ∞                          1                      (            6            n            +            1                          )              2                                                                        =                    −                  2          9                                      π            2                    8                +                              π            2                    9                                                  =                                          π            2                    12                .            Here      ∑          n      =      0        ∞        1          (      2      n      +      1              )        2              =            π      2        8    ,      ∑          n      =      −      ∞        ∞        1          (      6      n      +      1              )        2              =            π      2        9  are used.

sg101cp6vv · Accepted Answer

If we have   g  (  z  ) with a simple zero at    z  =      z    0  , then                    (1a)                              Res                      z            =                          z              0                                      ⁡                  (                                    f              (              z              )                                      g              (              z              )                                )                                    =                  lim                      z            →                          z              0                                                            (            z            −                          z              0                        )            f            (            z            )                                g            (            z            )                                              (1b)                                  =                              f            (                          z              0                        )                                              g              ′                        (                          z              0                        )                              Applying (1), we get                    (2a)                              Res                      z            =                          e                              π                i                                  /                                3                                                    ⁡                  (                                    z              2                                                      z                4                            +                              z                2                            +              1                                )                                    =                  1                      4            z            +            2                          /                        z                                              (2b)                                  =                              3            −            i                          3                                12                                        (3a)                              Res                      z            =                          e                              π                i                                  /                                3                                                    ⁡                  (                                                    z                2                                            log                ⁡                (                z                )                                                                    z                4                            +                              z                2                            +              1                                )                                    =                              3            −            i                          3                                12                                                    π              i                        3                                              (3b)                                  =        π                                            3                        +            3            i                    36                                        (4a)                              Res                      z            =                          e                              2                π                i                                  /                                3                                                    ⁡                  (                                    z              2                                                      z                4                            +                              z                2                            +              1                                )                                    =                  1                      4            z            +            2                          /                        z                                              (4b)                                  =                              −            3            −            i                          3                                12                                        (5a)                              Res                      z            =                          e                              2                π                i                                  /                                3                                                    ⁡                  (                                                    z                2                                            log                ⁡                (                z                )                                                                    z                4                            +                              z                2                            +              1                                )                                    =                              −            3            −            i                          3                                12                                                    2              π              i                        3                                              (5b)                                  =        π                              2                          3                        −            6            i                    36                    Applying the ResiduesYou had gotten                    (6)                              ∫                      −            ∞                    ∞                                                    z              2                        log            ⁡            (            z            )                                      d                        z                                              z              4                        +                          z              2                        +            1                          =        2                  ∫          0          ∞                                                    z              2                        log            ⁡            (            z            )                                      d                        z                                              z              4                        +                          z              2                        +            1                          +                              π            i                    2                          ∫                      −            ∞                    ∞                                                    z              2                                                  d                        z                                              z              4                        +                          z              2                        +            1                              which gives                    (7a)                              ∫          0          ∞                                                    z              2                        log            ⁡            (            z            )                                      d                        z                                              z              4                        +                          z              2                        +            1                                              =                  1          2                          ∫                      −            ∞                    ∞                                                    z              2                        log            ⁡            (            z            )                                      d                        z                                              z              4                        +                          z              2                        +            1                          −                              π            i                    4                          ∫                      −            ∞                    ∞                                                    z              2                                                  d                        z                                              z              4                        +                          z              2                        +            1                                              (7b)                                  =        π        i                  (          π                                                    3                            −              i                        12                    )                +                              π            2                    2                          (          −                                    i                              3                                      6                    )                                    (7c)                                  =                              π            2                    12                    Explanation:(7a): algebraic manipulation of (6)(7b): apply (3) and (5) to the first integral and (2) and (4) to the second integral(7c): evaluateWe used an upper half-plane semi-circular contour in both integrals on the right-hand side of (7a). Therefore, we included the residues from the poles at       e          π      i              /            3       and       e          2      π      i              /            3

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