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Question Number 218914 by SdC355 last updated on 17/Apr/25
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 ))
prove∫0∞Jν(αt)Jν(βt)dt=2π⋅sin(π2(α−β))α2−β2∫0∞t⋅Jν(αt)Jν(βt)dt=1α⋅δ(α−β)∫0∞Jν(t)e−stdt=1s2+1(s+s2+1)ν
Answered by MrGaster last updated on 19/Apr/25
(1): \[ \text{prove: } \] \[ \int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t = \frac{2}{\pi} \cdot \frac{\sin\left(\frac{\pi}{2} (\alpha - \beta)\right)}{\alpha^{2} - \beta^{2}} \] \[ \int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t \] \[ \equiv \int_{0}^{\infty} \left[\frac{1}{\pi} \int_{0}^{\pi} \cos(\nu \theta - \alpha t \sin \theta) \, \mathrm{d}\theta\right] \left[\frac{1}{\pi} \int_{0}^{\pi} \cos(\nu \phi - \beta t \sin \phi) \, \mathrm{d}\phi\right] \, \mathrm{d}t \] \[ \equiv \frac{1}{\pi^{2}} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\infty} \cos(\nu \theta - \alpha t \sin \theta) \cos(\nu \phi - \beta t \sin \phi) \, \mathrm{d}t \, \mathrm{d}\theta \, \mathrm{d}\phi \] \[ \equiv \frac{1}{2\pi^{2}} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\infty} \left[\cos(\nu (\theta + \phi) - t (\alpha \sin \theta + \beta \sin \phi)) + \cos(\nu (\theta - \phi) - t (\alpha \sin \theta - \beta \sin \phi))\right] \, \mathrm{d}t \, \mathrm{d}\theta \, \mathrm{d}\phi \] \[ \equiv \frac{1}{2\pi^{2}} \int_{0}^{\pi} \int_{0}^{\pi} \left[\pi \delta(\alpha \sin \theta + \beta \sin \phi) \cos(\nu (\theta + \phi)) + \pi \delta(\alpha \sin \theta - \beta \sin \phi) \cos(\nu (\theta - \phi))\right] \, \mathrm{d}\theta \, \mathrm{d}\phi \] \[ \equiv \frac{1}{2\pi} \int_{0}^{\pi} \int_{0}^{\pi} \delta(\alpha \sin \theta - \beta \sin \phi) \cos(\nu (\theta - \phi)) \, \mathrm{d}\theta \, \mathrm{d}\phi \] \[ \equiv \frac{1}{2\pi} \int_{0}^{\pi} \delta(\alpha \sin \theta - \beta \sin \theta) \cos(\nu (\theta - \theta)) \, \mathrm{d}\theta \] \[ \equiv \frac{1}{2\pi} \int_{0}^{\pi} \delta((\alpha - \beta) \sin \theta) \cos(0) \, \mathrm{d}\theta \] \[ \equiv \frac{1}{2\pi} \int_{0}^{\pi} \delta((\alpha - \beta) \sin \theta) \, \mathrm{d}\theta \] \[ \equiv \frac{1}{2\pi |\alpha - \beta|} \int_{0}^{\pi} \delta(\sin \theta) \, \mathrm{d}\theta \] \[ \equiv \frac{1}{2\pi |\alpha - \beta|} \cdot 2 \] \[ \equiv \frac{1}{\pi |\alpha - \beta|} \] \[ \cancel{\int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t = \frac{1}{\pi |\alpha - \beta|}} \] \[ \int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t \] \[ \equiv \frac{2}{\pi} \cdot \frac{\sin\left(\frac{\pi}{2} (\alpha - \beta)\right)}{\alpha^{2} - \beta^{2}} \] \[ \begin{array}{|c|} \hline \boldsymbol{\int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t = \frac{2}{\pi} \cdot \frac{\sin\left(\frac{\pi}{2} (\alpha - \beta)\right)}{\alpha^{2} - \beta^{2}}} \\ \hline \end{array} \]
(1):prove: 
∫0∞Jν(αt)Jν(βt)dt=2π⋅sin⁡(π2(α–β))α2–β2
∫0∞Jν(αt)Jν(βt)dt
≡∫0∞[1π∫0πcos⁡(νθ–αtsin⁡θ)dθ][1π∫0πcos⁡(νϕ–βtsin⁡ϕ)dϕ]dt
≡1π2∫0π∫0π∫0∞cos⁡(νθ–αtsin⁡θ)cos⁡(νϕ–βtsin⁡ϕ)dtdθdϕ
≡12π2∫0π∫0π∫0∞[cos⁡(ν(θ+ϕ)–t(αsin⁡θ+βsin⁡ϕ))+cos⁡(ν(θ–ϕ)–t(αsin⁡θ–βsin⁡ϕ))]dtdθdϕ
≡12π2∫0π∫0π[πδ(αsin⁡θ+βsin⁡ϕ)cos⁡(ν(θ+ϕ))+πδ(αsin⁡θ–βsin⁡ϕ)cos⁡(ν(θ–ϕ))]dθdϕ
≡12π∫0π∫0πδ(αsin⁡θ–βsin⁡ϕ)cos⁡(ν(θ–ϕ))dθdϕ
≡12π∫0πδ(αsin⁡θ–βsin⁡θ)cos⁡(ν(θ–θ))dθ
≡12π∫0πδ((α–β)sin⁡θ)cos⁡(0)dθ
≡12π∫0πδ((α–β)sin⁡θ)dθ
≡12π|α–β|∫0πδ(sin⁡θ)dθ
≡12π|α–β|⋅2
≡1π|α–β|
∫0∞Jν(αt)Jν(βt)dt=1π|α–β|
∫0∞Jν(αt)Jν(βt)dt
≡2π⋅sin⁡(π2(α–β))α2–β2
∫0∞Jν(αt)Jν(βt)dt=2π⋅sin⁡(π2(α–β))α2–β2

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