We know the Hankel transform of order 0 is defined as F 0 ( k ) = ∫ 0 ∞ f ( r ) J 0 ( k r ) r d r .In this regard, I am now trying to calculate the Hankel transform of the function f ( r ) = e − a r 2 + z 2 r 2 + z 2 ,with a ∈ R . Unfortunately, I was only able to obtain the solution for a = 0 . Any thoughts on how to solve this?

Question

We know the Hankel transform of order 0 is defined as                    F                  0                    (      k      )      =              ∫                  0                          ∞                    f      (      r      )              J                  0                    (      k      r      )            r                    d            r        .In this regard, I am now trying to calculate the Hankel transform of the function  f  (  r  )  =            e              −        a                              r            2                    +                      z            2                                              r        2            +              z        2              ,with   a  ∈      R   . Unfortunately, I was only able to obtain the solution for   a  =  0 . Any thoughts on how to solve this?

Paxton James · Accepted Answer

(1)                    F        (        k        ,        a        ,        z        )        :=                  ∫          0          ∞                                      exp            ⁡                                                            (                                                          −              a                                                r                  2                                +                                  z                  2                                                                              )                                                                                        r              2                        +                          z              2                                                J          0                (        k                 r        )                 r        d        r        =                              exp            ⁡                                                            (                                                          −              z                                                a                  2                                +                                  k                  2                                                                              )                                                                                        a              2                        +                          k              2                                                      exp      ⁡                                    (                                  −        a                              r            2                    +                      z            2                                                )                                      a                        r          2                +                  z          2                      =      ∫    1    ∞        e          −      a             r             t            J    0    (  a     z            t      2        −    1    )  d  tInsert into left-hand side of (1) and interchange   ∫  . The innermost integral has a closed form,      ∫    0    ∞        e          −      a             r             t            J    0    (  k     r  )  r  d  r  =            a             t                                (                    (      a      t              )        2            +              k        2                                          )                                    3                      /                    2                    which Mathematica knows. So we now have                    (2a)                    F        (        k        ,        a        ,        z        )        =                  a          2                          ∫          1          ∞                          J          0                (        a                 z                              t            2                    −          1                )                              t                         d            t                                                              (                                      (            a            t                          )              2                        +                          k              2                                                                        )                                                            3                                  /                                2                                                                            (2b)                    =                  1                      2            a                                    ∫          1          ∞                                                    J              0                        (            a                         z                          t              −              1                        )                                                              (                                      t            +            (            k                          /                        a                          )              2                                                                        )                                                            3                                  /                                2                                                    d        t        =                  1                      2            a                                    ∫          0          ∞                                                    J              0                        (            a                         z                          t                        )                                                              (                                      t            +            p                                                            )                                                            3                                  /                                2                                                    d        t         with         p        =        1        +        (        k                  /                a                  )          2                    where from line 2a to 2b we substituted       t    2    →  t and in the last step the limits of the integrand have been shifted with a subsequent change of the parameter. Now use the integral relationship      ∫    0    ∞        J    0    (  c      u    )  (  u  +  p      )          −      3              /            2        d  u  =  2            exp      ⁡                                    (                                  −        c                  p                                      )                                      p        .Mathematica knows this integral with   c  =  1  , and it is easy to work in the scaling factor. Algebra completes the proof.

We know the Hankel transform of order 0 is defined as <mrow class="MJX-TeXAtom-ORD"> <mstyle

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