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 ))


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Question Number 218914 by SdC355 last updated on 17/Apr/25

$prove$ $\int_{0}^{\infty} J_{ν} (α t) J_{ν} (β t) d t = \frac{2}{π} \cdot \frac{\sin (\frac{π}{2} (α - β))}{α^{2} - β^{2}}$ $\int_{0}^{\infty} t \cdot J_{ν} (α t) J_{ν} (β t) d t = \frac{1}{α} \cdot δ (α - β)$ $\int_{0}^{\infty} J_{ν} (t) e^{- s t} d t = \frac{1}{\sqrt{s^{2} + 1} {(s + \sqrt{s^{2} + 1})}^{ν}}$
	Answered by MrGaster last updated on 19/Apr/25
	$(1): \[ \text{prove: } \] \[ \int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t = \frac{2}{\pi} \cdot \frac{\sin\left(\frac{\pi}{2} (\alpha - \beta)\right)}{\alpha^{2} - \beta^{2}} \] \[ \int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t \] \[ \equiv \int_{0}^{\infty} \left[\frac{1}{\pi} \int_{0}^{\pi} \cos(\nu \theta - \alpha t \sin \theta) \, \mathrm{d}\theta\right] \left[\frac{1}{\pi} \int_{0}^{\pi} \cos(\nu \phi - \beta t \sin \phi) \, \mathrm{d}\phi\right] \, \mathrm{d}t \] \[ \equiv \frac{1}{\pi^{2}} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\infty} \cos(\nu \theta - \alpha t \sin \theta) \cos(\nu \phi - \beta t \sin \phi) \, \mathrm{d}t \, \mathrm{d}\theta \, \mathrm{d}\phi \] \[ \equiv \frac{1}{2\pi^{2}} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\infty} \left[\cos(\nu (\theta + \phi) - t (\alpha \sin \theta + \beta \sin \phi)) + \cos(\nu (\theta - \phi) - t (\alpha \sin \theta - \beta \sin \phi))\right] \, \mathrm{d}t \, \mathrm{d}\theta \, \mathrm{d}\phi \] \[ \equiv \frac{1}{2\pi^{2}} \int_{0}^{\pi} \int_{0}^{\pi} \left[\pi \delta(\alpha \sin \theta + \beta \sin \phi) \cos(\nu (\theta + \phi)) + \pi \delta(\alpha \sin \theta - \beta \sin \phi) \cos(\nu (\theta - \phi))\right] \, \mathrm{d}\theta \, \mathrm{d}\phi \] \[ \equiv \frac{1}{2\pi} \int_{0}^{\pi} \int_{0}^{\pi} \delta(\alpha \sin \theta - \beta \sin \phi) \cos(\nu (\theta - \phi)) \, \mathrm{d}\theta \, \mathrm{d}\phi \] \[ \equiv \frac{1}{2\pi} \int_{0}^{\pi} \delta(\alpha \sin \theta - \beta \sin \theta) \cos(\nu (\theta - \theta)) \, \mathrm{d}\theta \] \[ \equiv \frac{1}{2\pi} \int_{0}^{\pi} \delta((\alpha - \beta) \sin \theta) \cos(0) \, \mathrm{d}\theta \] \[ \equiv \frac{1}{2\pi} \int_{0}^{\pi} \delta((\alpha - \beta) \sin \theta) \, \mathrm{d}\theta \] \[ \equiv \frac{1}{2\pi \|\alpha - \beta\|} \int_{0}^{\pi} \delta(\sin \theta) \, \mathrm{d}\theta \] \[ \equiv \frac{1}{2\pi \|\alpha - \beta\|} \cdot 2 \] \[ \equiv \frac{1}{\pi \|\alpha - \beta\|} \] \[ \cancel{\int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t = \frac{1}{\pi \|\alpha - \beta\|}} \] \[ \int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t \] \[ \equiv \frac{2}{\pi} \cdot \frac{\sin\left(\frac{\pi}{2} (\alpha - \beta)\right)}{\alpha^{2} - \beta^{2}} \] \[ \begin{array}{\|c\|} \hline \boldsymbol{\int_{0}^{\infty} J_{\nu} (\alpha t) J_{\nu} (\beta t) \, \mathrm{d}t = \frac{2}{\pi} \cdot \frac{\sin\left(\frac{\pi}{2} (\alpha - \beta)\right)}{\alpha^{2} - \beta^{2}}} \\ \hline \end{array} \]$
	$(1) :$ $prove:$ $\int_{0}^{\infty} J_{ν} (α t) J_{ν} (β t) d t = \frac{2}{π} \cdot \frac{\sin (\frac{π}{2} (α - β))}{α^{2} - β^{2}}$ $\int_{0}^{\infty} J_{ν} (α t) J_{ν} (β t) d t$ $\equiv \int_{0}^{\infty} [\frac{1}{π} \int_{0}^{π} \cos (ν θ - α t \sin θ) d θ] [\frac{1}{π} \int_{0}^{π} \cos (ν ϕ - β t \sin ϕ) d ϕ] d t$ $\equiv \frac{1}{π^{2}} \int_{0}^{π} \int_{0}^{π} \int_{0}^{\infty} \cos (ν θ - α t \sin θ) \cos (ν ϕ - β t \sin ϕ) d t d θ d ϕ$ $\equiv \frac{1}{2 π^{2}} \int_{0}^{π} \int_{0}^{π} \int_{0}^{\infty} [\cos (ν (θ + ϕ) - t (α \sin θ + β \sin ϕ)) + \cos (ν (θ - ϕ) - t (α \sin θ - β \sin ϕ))] d t d θ d ϕ$ $\equiv \frac{1}{2 π^{2}} \int_{0}^{π} \int_{0}^{π} [π δ (α \sin θ + β \sin ϕ) \cos (ν (θ + ϕ)) + π δ (α \sin θ - β \sin ϕ) \cos (ν (θ - ϕ))] d θ d ϕ$ $\equiv \frac{1}{2 π} \int_{0}^{π} \int_{0}^{π} δ (α \sin θ - β \sin ϕ) \cos (ν (θ - ϕ)) d θ d ϕ$ $\equiv \frac{1}{2 π} \int_{0}^{π} δ (α \sin θ - β \sin θ) \cos (ν (θ - θ)) d θ$ $\equiv \frac{1}{2 π} \int_{0}^{π} δ ((α - β) \sin θ) \cos (0) d θ$ $\equiv \frac{1}{2 π} \int_{0}^{π} δ ((α - β) \sin θ) d θ$ $\equiv \frac{1}{2 π \| α - β \|} \int_{0}^{π} δ (\sin θ) d θ$ $\equiv \frac{1}{2 π \| α - β \|} \cdot 2$ $\equiv \frac{1}{π \| α - β \|}$ $\int_{0}^{\infty} J_{ν} (α t) J_{ν} (β t) d t = \frac{1}{π \| α - β \|}$ $\int_{0}^{\infty} J_{ν} (α t) J_{ν} (β t) d t$ $\equiv \frac{2}{π} \cdot \frac{\sin (\frac{π}{2} (α - β))}{α^{2} - β^{2}}$ $\begin{matrix} \int_{0}^{\infty} J_{ν} (α t) J_{ν} (β t) d t = \frac{2}{π} \cdot \frac{\sin (\frac{π}{2} (α - β))}{α^{2} - β^{2}} \end{matrix}$
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