Consider the integral I ( x ) = ∫ 0 ∞ exp ⁡ ( − s 2 / 2 ) cos ⁡ ( x s + λ s 3 ) d s ,

Question

Consider the integral  I  (  x  )  =      ∫          0              ∞        exp  ⁡            (        −      s    2        /    2            )        cos  ⁡            (        x  s  +  λ      s    3              )          d  s  ,

kejohananws · Accepted Answer

This is an elaboration on Maxim&#039;s comment. Your integral is  I  (  x  )  =      1    2        ∫          −      ∞              +      ∞            exp    ⁡          (              −                  1          2                          s          2                +        i        λ                  s          3                +        i        x        s            )        d    s    .With the change of variables   s  =      t          (      3      λ              )                  1                      /                    3                      +      1          6      i      λ      , we find  I  (  x  )    =        1    2        1          (      3      λ              )                  1                      /                    3                      exp  ⁡      (                  x                  6          λ                          +                    1                  108                      λ            2                                )        ∫          −      ∞      +      i              /            (      6      λ      )              +      ∞      +      i              /            (      6      λ      )            exp    ⁡          (              i                  (                                                                      t                  3                                            3                                    +                                      1                              (                3                λ                                  )                                      1                                          /                                        3                                                                                                  (                              x                                +                                                  1                                      12                    λ                                                              )                        t                    )                    )        d    t    .Pushing the contour downwards and using the know integral representation of the Airy function, we deduce                    I        (        x        )                            =                  1          2                          1                      (            3            λ                          )                              1                                  /                                3                                                    exp        ⁡                  (                                    x                              6                λ                                      +                          1                              108                                  λ                  2                                                              )                          ∫                      −            ∞                                +            ∞                                    exp          ⁡                      (                          i                              (                                                                                                    t                        3                                                              3                                    +                                      1                                          (                      3                      λ                                              )                                                  1                                                      /                                                    3                                                                                                                          (                                          x                      +                                              1                                                  12                          λ                                                                                      )                                    t                                )                                      )                    d          t                                                  =                  1                      (            3            λ                          )                              1                                  /                                3                                                    exp        ⁡                  (                                    x                              6                λ                                      +                          1                              108                                  λ                  2                                                              )                          ∫                      −            ∞                                +            ∞                                    cos          ⁡                      (                                                                                t                    3                                                  3                            +                              1                                  (                  3                  λ                                      )                                          1                                              /                                            3                                                                                                  (                                  x                  +                                      1                                          12                      λ                                                                      )                            t                        )                    d          t                                                  =                  π                      (            3            λ                          )                              1                                  /                                3                                                    exp        ⁡                  (                                    x                              6                λ                                      +                          1                              108                                  λ                  2                                                              )                Ai        ⁡                  (                      1                          (              3              λ                              )                                  1                                      /                                    3                                                                          (                          x              +                              1                                  12                  λ                                                      )                    )                .            Now we know that  Ai  ⁡  (  z  )  =      1          2              π                    z                  1                      /                    4                      exp  ⁡      (          −                        2                      z                          3                              /                            2                                      3              )        (          1      +              O                          (                              1                                          z                                  3                                      /                                    2                                                                    )              )  as   z  →  ∞ in the sector       |    arg  ⁡  z      |    ≤  π  −  δ). Since      2    3        1                  3        λ                        (              x        +                  1                      12            λ                              )              3              /            2        =            2              x                  3                      /                    2                            3              3        λ              +                    x                  1                      /                    2                            12              3                    λ                  3                      /                    2                      +      1          576              3                    λ                  5                      /                    2                            x                  1                      /                    2                      −  ⋯and  (  3  λ      )          1              /            12                  (              x        +                  1                      12            λ                              )              −      1              /            4        =  (  3  λ      )          1              /            12            x          −      1              /            4            (          1      −              1                  48          λ          x                    +      ⋯        )    ,we obtain  I  (  x  )  =                    π                    2      (      3      λ      x              )                  1                      /                    4                      exp  ⁡      (          −                        2                      x                          3                              /                            2                                                3                      3            λ                              +              x                  6          λ                    −                                    x                          1                              /                            2                                                12                      3                                λ                          3                              /                            2                                          +              1                  108                      λ            2                                )        (          1      +              O                          (                              1                                          x                                  1                                      /                                    2                                                                    )              )  as   x  →  ∞ in the sector       |    arg  ⁡  x      |    ≤  π  −  δ). More terms in the expansion can be obtained if necessary.

Consider the integral I ( x ) = <msubsup> ∫ <mrow class="MJX-

Answered question

Answer & Explanation

New Questions in Integral Calculus