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Question Number 218890    Answers: 5   Comments: 0

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Question Number 218879    Answers: 0   Comments: 0

Calculate the following integral; ∫_(−∞) ^∞ ∫_(−∞) ^∞ ∫_(−∞) ^∞ xJ_0 ((√(x^2 +y^2 )))J_1 ((√(y^2 +z^2 )))J_2 ((√(z^2 +x^2 )))e^(−(x^2 +y^2 +z^2 )) dxdydz where J_n (u) is the Bassel function of the first kind of order n

$$ \\ $$$$\:\:\:\boldsymbol{{Calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{following}}\:\boldsymbol{{integral}};\:\:\:\:\:\: \\ $$$$\:\:\int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} \int_{−\infty} ^{\infty} \boldsymbol{{xJ}}_{\mathrm{0}} \left(\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }\right)\boldsymbol{{J}}_{\mathrm{1}} \left(\sqrt{\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} }\right)\boldsymbol{{J}}_{\mathrm{2}} \left(\sqrt{\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} }\right)\boldsymbol{{e}}^{−\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} \right)} \boldsymbol{{dxdydz}}\:\:\:\:\:\:\: \\ $$$$\:\:\:\boldsymbol{{where}}\:\boldsymbol{{J}}_{\boldsymbol{{n}}} \left(\boldsymbol{{u}}\right)\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{Bassel}}\:\boldsymbol{{function}} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{first}}\:\boldsymbol{{kind}}\:\boldsymbol{{of}}\:\boldsymbol{{order}}\:\boldsymbol{{n}} \\ $$$$ \\ $$

Question Number 218872    Answers: 0   Comments: 0

Calculate the following integral; ∫_0 ^∞ ∫_0 ^∞ ∫_0 ^∞ ((J_𝛂 (ax)J_𝛃 (by)J_𝛄 (cz))/( (√(x^2 +y^2 +z^2 )))) e^(−p(x^2 +y^2 +z^2 ) dxdydz ) where; • J_ν (u) is the Bassel function of the first kind of order 𝛎. • 𝛂,𝛃 and 𝛄 are arbitrary real numbers parameters (Not necessarily integers) • a,b,c and p are positive constants

$$ \\ $$$$\:\boldsymbol{{Calculate}}\:\boldsymbol{{the}}\:\boldsymbol{{following}}\:\boldsymbol{{integral}}; \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:\frac{\boldsymbol{{J}}_{\boldsymbol{\alpha}} \left(\boldsymbol{{ax}}\right)\boldsymbol{{J}}_{\boldsymbol{\beta}} \left(\boldsymbol{{by}}\right)\boldsymbol{{J}}_{\boldsymbol{\gamma}} \left(\boldsymbol{{cz}}\right)}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} }}\:\boldsymbol{{e}}^{−\boldsymbol{{p}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} \right)\:\boldsymbol{{dxdydz}}\:\:\:\:\:} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{where}}; \\ $$$$\:\:\:\:\bullet\:\:\boldsymbol{{J}}_{\nu} \left(\boldsymbol{{u}}\right)\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{Bassel}}\:\boldsymbol{{function}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{first}}\:\boldsymbol{{kind}}\:\boldsymbol{{of}}\:\boldsymbol{{order}}\:\boldsymbol{\nu}.\:\:\:\:\: \\ $$$$\:\:\:\:\bullet\:\:\boldsymbol{\alpha},\boldsymbol{\beta}\:{and}\:\boldsymbol{\gamma}\:\boldsymbol{{are}}\:\boldsymbol{{arbitrary}}\:\boldsymbol{{real}}\:\boldsymbol{{numbers}}\:\boldsymbol{{parameters}}\: \\ $$$$\:\:\:\:\:\:\:\:\left(\boldsymbol{{Not}}\:\boldsymbol{{necessarily}}\:\boldsymbol{{integers}}\right) \\ $$$$\:\:\:\:\bullet\:\:\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}\:\boldsymbol{{and}}\:\boldsymbol{{p}}\:\boldsymbol{{are}}\:\boldsymbol{{positive}}\:\boldsymbol{{constants}}\:\:\:\: \\ $$$$ \\ $$

Question Number 218914    Answers: 1   Comments: 0

prove ∫_0 ^( ∞) J_ν (αt)J_ν (βt)dt=(2/π)∙((sin((π/2)(α−β)))/(α^2 −β^2 )) ∫_0 ^( ∞) t∙J_ν (αt)J_ν (βt)dt=(1/α)∙δ(α−β) ∫_0 ^( ∞) J_ν (t)e^(−st) dt=(1/( (√(s^2 +1))(s+(√(s^2 +1)))^ν ))

$$\mathrm{prove} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{\nu} \left(\alpha{t}\right){J}_{\nu} \left(\beta{t}\right)\mathrm{d}{t}=\frac{\mathrm{2}}{\pi}\centerdot\frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\left(\alpha−\beta\right)\right)}{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{t}\centerdot{J}_{\nu} \left(\alpha{t}\right){J}_{\nu} \left(\beta{t}\right)\mathrm{d}{t}=\frac{\mathrm{1}}{\alpha}\centerdot\delta\left(\alpha−\beta\right) \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{\nu} \left({t}\right){e}^{−{st}} \mathrm{d}{t}=\frac{\mathrm{1}}{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{\nu} } \\ $$

Question Number 218907    Answers: 1   Comments: 2

_0 ∫^( 45) arctan(((1+tan x)/( (√2))))dx = ?

$$\:_{\mathrm{0}} \int^{\:\mathrm{45}} {arctan}\left(\frac{\mathrm{1}+{tan}\:{x}}{\:\sqrt{\mathrm{2}}}\right){dx}\:=\:? \\ $$

Question Number 218866    Answers: 0   Comments: 0

evaluate the following integral in closed form or express it in terms of known special functions; ∫_ ^∞ K_(i𝛌) (at)J_𝛎 (bt)^(𝛍−1) dt where; •K_(i𝛌) (z) is the modified Bessel function of the second kind with imaginary order i𝛌, where 𝛌∈R. • J_𝛎 (z) is the Bessel function of the first kind with order 𝛎 ∈C. • a,b are positif real constants. • 𝛍 is a complex parameter statisfying the conditions for the integral to converge.

$$ \\ $$$$\:\:\:{evaluate}\:{the}\:{following}\:{integral}\:{in}\:{closed}\:{form}\:{or}\:{express} \\ $$$$\:{it}\:{in}\:{terms}\:{of}\:{known}\:{special}\:{functions};\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\int_{ } ^{\infty} \boldsymbol{{K}}_{\boldsymbol{{i}\lambda}} \left(\boldsymbol{{at}}\right)\boldsymbol{{J}}_{\boldsymbol{\nu}} \left(\boldsymbol{{bt}}\right)^{\boldsymbol{\mu}−\mathrm{1}} \boldsymbol{{dt}} \\ $$$$\:\boldsymbol{{where}}; \\ $$$$\:\:\bullet\boldsymbol{{K}}_{\boldsymbol{{i}\lambda}} \left(\boldsymbol{{z}}\right)\:{is}\:{the}\:{modified}\:\boldsymbol{{B}}{essel}\:{function}\: \\ $$$$\:\:\:\:\:\:{of}\:{the}\:{second}\:{kind}\:{with}\:{imaginary}\:{order}\:\boldsymbol{{i}\lambda}, \\ $$$$\:\:\:\:\:\:\:{where}\:\boldsymbol{\lambda}\in\mathbb{R}. \\ $$$$\:\:\bullet\:\boldsymbol{{J}}_{\boldsymbol{\nu}} \left(\boldsymbol{{z}}\right)\:{is}\:{the}\:{Bessel}\:{function}\:{of}\:{the}\:{first}\:{kind}\:{with}\:{order}\:\boldsymbol{\nu}\:\in\mathbb{C}.\:\:\:\:\: \\ $$$$\:\:\bullet\:\boldsymbol{{a}},\boldsymbol{{b}}\:{are}\:{positif}\:{real}\:{constants}. \\ $$$$\:\:\bullet\:\boldsymbol{\mu}\:{is}\:{a}\:{complex}\:{parameter}\:{statisfying}\:{the}\:\:\: \\ $$$$\:\:\:\:\:\:\:{conditions}\:{for}\:{the}\:{integral}\:{to}\:{converge}.\:\:\:\: \\ $$$$ \\ $$$$\:\:\: \\ $$

Question Number 218857    Answers: 0   Comments: 0

((d )/dt) ((dx(t))/dt)−(x(t))^2 =k_0 ^2 ...?? how can i solve this Differantial Equation...???

$$\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\:\frac{\mathrm{d}{x}\left({t}\right)}{\mathrm{d}{t}}−\left({x}\left({t}\right)\right)^{\mathrm{2}} ={k}_{\mathrm{0}} ^{\mathrm{2}} ...?? \\ $$$$\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{Differantial}\:\mathrm{Equation}...??? \\ $$

Question Number 218855    Answers: 0   Comments: 0

(√3)x^2 =(√((64−x^2 )(x^2 −4)))+(√((81−x^2 )(x^2 −1)))+(√((49−x^2 )(x^2 −1)))solve

$$\sqrt{\mathrm{3}}{x}^{\mathrm{2}} =\sqrt{\left(\mathrm{64}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{4}\right)}+\sqrt{\left(\mathrm{81}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)}+\sqrt{\left(\mathrm{49}−{x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\boldsymbol{{solve}} \\ $$

Question Number 218853    Answers: 5   Comments: 0

Question Number 218854    Answers: 3   Comments: 0

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Question Number 218975    Answers: 0   Comments: 0

In physics , Flux integral ∮_( ∂S) F^→ ∙ dS^→ is a concept that widely used in eletric equation or Heat Eqaution for example..... ∮_( A) D^→ ∙dA^→ =Q_0 (Gauss law) D^→ is displayment field ∮_( S) B^→ ∙dA^→ =0 (Gauss law for magnetic) B^→ is Magnetic field and in Heat Flux (∂E_(in) /∂t)−(∂E_(out) /∂t)=∮_( S) 𝛗_q ^→ ∙dS^→ But in mathematic it seems that Surface integral in the vector field is only extended version of the integral,why mathematic don′t use surface integral like physics...??? i really curious

$$\mathrm{In}\:\mathrm{physics}\:,\:\mathrm{Flux}\:\mathrm{integral}\:\oint_{\:\partial\boldsymbol{\mathcal{S}}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\:\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathrm{S}}}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{concept}\:\mathrm{that}\:\mathrm{widely}\:\mathrm{used}\:\mathrm{in}\:\mathrm{eletric}\:\mathrm{equation}\:\mathrm{or} \\ $$$$\mathrm{Heat}\:\mathrm{Eqaution} \\ $$$$\mathrm{for}\:\mathrm{example}.....\: \\ $$$$\oint_{\:{A}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{D}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathrm{A}}}={Q}_{\mathrm{0}} \:\left(\mathrm{Gauss}\:\mathrm{law}\right)\:\overset{\rightarrow} {\boldsymbol{\mathrm{D}}}\:\mathrm{is}\:\mathrm{displayment}\:\mathrm{field} \\ $$$$\oint_{\:{S}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{B}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathrm{A}}}=\mathrm{0}\:\left(\mathrm{Gauss}\:\mathrm{law}\:\mathrm{for}\:\mathrm{magnetic}\right)\:\overset{\rightarrow} {\boldsymbol{\mathrm{B}}}\:\mathrm{is}\:\mathrm{Magnetic}\:\mathrm{field} \\ $$$$\mathrm{and}\:\mathrm{in}\:\mathrm{Heat}\:\mathrm{Flux} \\ $$$$\frac{\partial\mathrm{E}_{\mathrm{in}} }{\partial{t}}−\frac{\partial\mathrm{E}_{\mathrm{out}} }{\partial{t}}=\oint_{\:{S}} \:\overset{\rightarrow} {\boldsymbol{\phi}}_{\mathrm{q}} \centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathrm{S}}}\:\: \\ $$$$\mathrm{But}\:\mathrm{in}\:\mathrm{mathematic}\:\mathrm{it}\:\mathrm{seems}\:\mathrm{that}\:\mathrm{Surface}\:\mathrm{integral} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{vector}\:\mathrm{field}\:\mathrm{is}\:\mathrm{only}\:\mathrm{extended}\:\mathrm{version}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{integral},\mathrm{why}\:\mathrm{mathematic}\:\mathrm{don}'\mathrm{t}\:\mathrm{use}\:\mathrm{surface}\:\mathrm{integral} \\ $$$$\mathrm{like}\:\mathrm{physics}...???\:\mathrm{i}\:\mathrm{really}\:\mathrm{curious}\: \\ $$

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