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Question Number 212765 by issac last updated on 23/Oct/24

i generalized Bessel {function}^{'} s

Laplace Transform

L . T J_{ν} (z) = \frac{{(s + \sqrt{s^{2} + 1})}^{- ν}}{\sqrt{s^{2} + 1}}, s \in [0, \infty), ν \in R

L . T Y_{ν} (z) = \frac{\cot (π ν) {(s + \sqrt{s^{2} + 1})}^{- ν}}{\sqrt{s^{2} + 1}} - \frac{\csc (π ν) {(s + \sqrt{s^{2} + 1})}^{ν}}{\sqrt{s^{2} + 1}}

s \in [0, \infty), ν \in R^{+} ∖ {0, Z^{+}}

but \dots . i {can}^{'} t explain why L . T Y_{ν} (z)

is undefined when ν \in Z^{+} ∖ {0} \dots . .

Help me \dots . .!!!

Answered by Frix last updated on 23/Oct/24

\cot π ν = \frac{1}{\tan π ν} \Rightarrow \tan π ν \neq 0 \Rightarrow ν \neq n \forall n \in Z

\csc π ν = \frac{1}{\sin π ν} \Rightarrow \sin π ν \neq 0 \Rightarrow ν \neq n \forall n \in Z

Commented by Ghisom last updated on 24/Oct/24

let α \in Z, f_{α} (x) = \frac{1}{(1 - α^{2}) (1 - x^{2})} \Rightarrow f_{α} (x)

is not defind for x = \pm 1 or α = \pm . this means,

f_{α} (x) does exist for α \neq \pm 1 but with α = \pm 1

the function is not defined anywhere :

f_{- 1} (x) = f_{1} (x) = \frac{1}{0} . you cannot even say

f_{\pm 1} (x) = \pm \infty because {there}^{'} s no limit for

α \in Z

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