We know the Hankel transform of order 0 is defined as F 0 ( k...

dourtuntellorvl

dourtuntellorvl

Answered

2022-06-30

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?

Answer & Explanation

Paxton James

Paxton James

Expert

2022-07-01Added 25 answers

(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 t
Insert 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.

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