I have the integral I ( a ) = <munderover> &#x222B;<!-- ∫ --> 0 <mi mathva

Aganippe76 2022-07-06 Answered
I have the integral
I ( a ) = 0 d k   k 3 J 1 ( a k ) sinh ( k )
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Answers (1)

conveneau71
Answered 2022-07-07 Author has 17 answers
Well, we are trying to find the following integral:
(1) I n ( α ) := 0 x n J 1 ( α x ) sinh ( x )   d x = 0 x n J 1 ( α x ) csch ( x )   d x
Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:
(2) I n ( α ) = 0 L x [ x n J 1 ( α x ) ] ( σ ) L x 1 [ csch ( x ) ] ( σ )   d σ
Using known results, we can write:
(3) I n ( α ) = ( 1 ) n 0 n σ n ( α α 2 + σ ( σ + α 2 + σ 2 ) ) ( 2 k     0 δ ( σ 2 k 1 ) )   d σ
We can rewrite this a bit:
(4) I n ( α ) = 2 ( 1 ) n k     0 0 n σ n ( α α 2 + σ ( σ + α 2 + σ 2 ) ) δ ( σ 2 k 1 )   d σ
Using the property:
(5) 0 y ( x ) δ ( x p )   d x = y ( n ) θ ( n )
We can write:
(6) I n ( α ) = 2 ( 1 ) n k     0 { θ ( 1 + 2 k ) n σ n ( α α 2 + σ ( σ + α 2 + σ 2 ) ) |   σ   =   1 + 2 k }
Now, using the fact that when k N we get θ ( 1 + 2 k ) = 1. So we can conclude:
(7) I n ( α ) = 2 ( 1 ) n k     0 { n σ n ( α α 2 + σ ( σ + α 2 + σ 2 ) ) |   σ   =   1 + 2 k }
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