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Question Number 213790 by issac last updated on 16/Nov/24
So Weird...... ∫_0 ^( ∞) J_ν (t)e^(−st) dt=(((s+(√(s^2 +1)))^(−ν) )/( (√(s^2 +1)))) J_(−ν) (t)=(−1)^ν J_ν (t) ∫_0 ^( ∞) J_(−ν) (t)e^(−st) dt=(((−1)^ν (s+(√(s^2 +1)))^(−ν) )/( (√(s^2 +1)))) is true But ∫_0 ^( ∞) J_(−ν) (t)e^(−st) dt is not (((s+(√(s^2 +1)))^ν )/( (√(s^2 +1)))) why....? can you explain why Blue equation is not true....
$$\mathrm{So}\:\mathrm{Weird}…… \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {J}_{\nu} \left({t}\right){e}^{−{st}} \mathrm{d}{t}=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\: \\ $$$${J}_{−\nu} \left({t}\right)=\left(−\mathrm{1}\right)^{\nu} {J}_{\nu} \left({t}\right)\:\: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{−\nu} \left({t}\right){e}^{−{st}} \mathrm{d}{t}=\frac{\left(−\mathrm{1}\right)^{\nu} \left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{is}\:\mathrm{true} \\ $$$$\mathrm{But}\:\int_{\mathrm{0}} ^{\:\infty} \:{J}_{−\nu} \left({t}\right){e}^{−{st}} \mathrm{d}{t}\:\mathrm{is}\:\mathrm{not}\:\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{why}….?\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\: \\ $$$$\mathrm{why}\:\mathrm{Blue}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{not}\:\mathrm{true}…. \\ $$
Commented by Frix last updated on 16/Nov/24
I know nothing about these functions but this is obvious, we only need this general rule of integration: ∫af(x)dx=a∫f(x)dx 1. ∫_0 ^∞ J_v (t)e^(−st) dt=(((s+(√(s^2 +1)))^(−v) )/( (√(s^2 +1)))) 2. J_(−v) (t)=(−1)^v J_v (t) ⇒ ∫_0 ^∞ J_(−v) (t)e^(−st) dt=(−1)^v ∫_0 ^∞ J_v (t)e^(−st) dt= =(((−1)^v (s+(√(s^2 +1)))^(−v) )/( (√(s^2 +1))))≠(((s+(√(s^2 +1)))^v )/( (√(s^2 +1)))) I don′t see the problem...
$$\mathrm{I}\:\mathrm{know}\:\mathrm{nothing}\:\mathrm{about}\:\mathrm{these}\:\mathrm{functions}\:\mathrm{but} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{obvious},\:\mathrm{we}\:\mathrm{only}\:\mathrm{need}\:\mathrm{this}\:\mathrm{general} \\ $$$$\mathrm{rule}\:\mathrm{of}\:\mathrm{integration}:\:\int{af}\left({x}\right){dx}={a}\int{f}\left({x}\right){dx} \\ $$$$\mathrm{1}.\:\underset{\mathrm{0}} {\overset{\infty} {\int}}{J}_{{v}} \left({t}\right)\mathrm{e}^{−{st}} {dt}=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−{v}} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{2}.\:{J}_{−{v}} \left({t}\right)=\left(−\mathrm{1}\right)^{{v}} {J}_{{v}} \left({t}\right) \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}{J}_{−{v}} \left({t}\right)\mathrm{e}^{−{st}} {dt}=\left(−\mathrm{1}\right)^{{v}} \underset{\mathrm{0}} {\overset{\infty} {\int}}{J}_{{v}} \left({t}\right)\mathrm{e}^{−{st}} {dt}= \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{v}} \left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−{v}} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\neq\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{{v}} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:\mathrm{the}\:\mathrm{problem}… \\ $$
Commented by issac last updated on 17/Nov/24
J_ν (z) is Σ_(h=0) ^∞ (((−1)^h )/(h!(h+ν)!)) ((z/2))^(2h+ν) Y_ν (z)=cot(πν)J_ν (z)−csc(πν)J_(−ν) (z) (Bessel function) Bessel function have some Properties ex. ν∈±2Z , J_(−ν) (z),Y_(−ν) (z)=J_ν (z),Y_ν (z) ν∉±2Z, J_(−ν) (z),Y_(−ν) (z)=−J_ν (z),−Y_ν (z) and ν∈Z J_(−ν−(1/2)) (z)=(−1)^(ν+1) Y_(ν+(1/2)) (z) Y_(−ν−(1/2)) (z)=(−1)^ν J_(ν+(1/2)) (z)
$${J}_{\nu} \left({z}\right)\:\mathrm{is}\:\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{h}} }{{h}!\left({h}+\nu\right)!}\:\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{h}+\nu} \\ $$$${Y}_{\nu} \left({z}\right)=\mathrm{cot}\left(\pi\nu\right){J}_{\nu} \left({z}\right)−\mathrm{csc}\left(\pi\nu\right){J}_{−\nu} \left({z}\right) \\ $$$$\left(\boldsymbol{\mathrm{Bessel}}\:\boldsymbol{\mathrm{function}}\right) \\ $$$$\boldsymbol{\mathrm{Bessel}}\:\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{some}}\:\boldsymbol{\mathrm{Properties}} \\ $$$$\boldsymbol{\mathrm{ex}}. \\ $$$$\nu\in\pm\mathrm{2}\mathbb{Z}\:,\:{J}_{−\nu} \left({z}\right),{Y}_{−\nu} \left({z}\right)={J}_{\nu} \left({z}\right),{Y}_{\nu} \left({z}\right) \\ $$$$\nu\notin\pm\mathrm{2}\mathbb{Z},\:{J}_{−\nu} \left({z}\right),{Y}_{−\nu} \left({z}\right)=−{J}_{\nu} \left({z}\right),−{Y}_{\nu} \left({z}\right) \\ $$$$\mathrm{and}\:\nu\in\mathbb{Z} \\ $$$${J}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)=\left(−\mathrm{1}\right)^{\nu+\mathrm{1}} {Y}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right) \\ $$$${Y}_{−\nu−\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right)=\left(−\mathrm{1}\right)^{\nu} {J}_{\nu+\frac{\mathrm{1}}{\mathrm{2}}} \left({z}\right) \\ $$

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