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Question Number 212765 by issac last updated on 23/Oct/24
i generalized Bessel function′s Laplace Transform L.TJ_ν (z)=(((s+(√(s^2 +1)))^(−ν) )/( (√(s^2 +1)))) , s∈[0,∞) , ν∈R L.T Y_ν (z)=((cot(πν)(s+(√(s^2 +1)))^(−ν) )/( (√(s^2 +1))))−((csc(πν)(s+(√(s^2 +1)))^ν )/( (√(s^2 +1)))) s∈[0,∞) , ν∈R^+ \{0,Z^+ } but.... i can′t explain why L.T Y_ν (z) is undefined when ν∈Z^+ \{0}..... Help me.....!!!
$$\mathrm{i}\:\:\mathrm{generalized}\:\boldsymbol{\mathrm{Bessel}}\:\boldsymbol{\mathrm{function}}'\mathrm{s} \\ $$$$\mathrm{Laplace}\:\mathrm{Transform} \\ $$$$\mathrm{L}.\mathrm{T}{J}_{\nu} \left({z}\right)=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\:,\:{s}\in\left[\mathrm{0},\infty\right)\:,\:\nu\in\mathbb{R} \\ $$$$\mathrm{L}.\mathrm{T}\:{Y}_{\nu} \left({z}\right)=\frac{\mathrm{cot}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}−\frac{\mathrm{csc}\left(\pi\nu\right)\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${s}\in\left[\mathrm{0},\infty\right)\:,\:\nu\in\mathbb{R}^{+} \backslash\left\{\mathrm{0},\mathbb{Z}^{+} \right\} \\ $$$$\mathrm{but}….\:\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{explain}\:\mathrm{why}\:\mathrm{L}.\mathrm{T}\:{Y}_{\nu} \left({z}\right) \\ $$$$\mathrm{is}\:\mathrm{undefined}\:\mathrm{when}\:\nu\in\mathbb{Z}^{+} \backslash\left\{\mathrm{0}\right\}….. \\ $$$$\mathrm{Help}\:\mathrm{me}…..!!! \\ $$
Answered by Frix last updated on 23/Oct/24
cot πν =(1/(tan πν)) ⇒ tan πν ≠0 ⇒ ν≠n∀n∈Z csc πν =(1/(sin πν)) ⇒ sin πν ≠0 ⇒ ν≠n∀n∈Z
$$ \\ $$$$\mathrm{cot}\:\pi\nu\:=\frac{\mathrm{1}}{\mathrm{tan}\:\pi\nu}\:\Rightarrow\:\mathrm{tan}\:\pi\nu\:\neq\mathrm{0}\:\Rightarrow\:\nu\neq{n}\forall{n}\in\mathbb{Z} \\ $$$$\mathrm{csc}\:\pi\nu\:=\frac{\mathrm{1}}{\mathrm{sin}\:\pi\nu}\:\Rightarrow\:\mathrm{sin}\:\pi\nu\:\neq\mathrm{0}\:\Rightarrow\:\nu\neq{n}\forall{n}\in\mathbb{Z} \\ $$
Commented by Ghisom last updated on 24/Oct/24
let α∈Z, f_α (x)=(1/((1−α^2 )(1−x^2 ))) ⇒ f_α (x) is not defind for x=±1 or α=±. this means, f_α (x) does exist for α≠±1 but with α=±1 the function is not defined anywhere: f_(−1) (x)=f_1 (x)=(1/0). you cannot even say f_(±1) (x)=±∞ because there′s no limit for α∈Z
$$\mathrm{let}\:\alpha\in\mathbb{Z},\:\:{f}_{\alpha} \left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}−\alpha^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:\Rightarrow\:{f}_{\alpha} \left({x}\right) \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{defind}\:\mathrm{for}\:{x}=\pm\mathrm{1}\:\mathrm{or}\:\alpha=\pm.\:\mathrm{this}\:\mathrm{means}, \\ $$$${f}_{\alpha} \left({x}\right)\:\mathrm{does}\:\mathrm{exist}\:\mathrm{for}\:\alpha\neq\pm\mathrm{1}\:\mathrm{but}\:\mathrm{with}\:\alpha=\pm\mathrm{1} \\ $$$$\mathrm{the}\:\:\mathrm{function}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{anywhere}: \\ $$$${f}_{−\mathrm{1}} \left({x}\right)={f}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{0}}.\:\mathrm{you}\:\mathrm{cannot}\:\mathrm{even}\:\mathrm{say} \\ $$$${f}_{\pm\mathrm{1}} \left({x}\right)=\pm\infty\:\mathrm{because}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{limit}\:\mathrm{for} \\ $$$$\alpha\in\mathbb{Z} \\ $$

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