Question Number 222805 by MrGaster last updated on 08/Jul/25
$$\mathrm{Prove}:\int_{−\infty} ^{\infty} {J}_{\mathrm{0}} \left(\mathrm{2}{x}\right){dx}=\mathrm{1} \\ $$
Answered by MrGaster last updated on 08/Jul/25
![J_0 (2x)=(1/(2π))∫_(−π) ^π e^(i2x sin φ) dφ ∫_(−∞) ^∞ J_0 (2x)dx=∫_(−∞) ^∞ (1/(2π))∫_(−π) ^π e^(i2x sin φ) dφdx =∫_(−π) ^π ∫_(−∞) ^∞ e^(i2x sin φ) dφdx ∫_(−∞) ^∞ e^(ikx) dx=2πkδ(k),k=2 sin φ ∫_(−∞) ^∞ e^(i2x sin φ) dx=2πδ(2 sin φ) δ(2 sin φ)=(1/2)δ(sin φ) =(1/(2π))∫_(−π) ^π 2π∙(1/2)δ(sin φ)dφ=∫_(−π) ^π (1/2)δ(sin φ)dφ φ∈[−π,π],sin φ=0 at φ=0,±π φ=−π≡π(periodicity) zeros:θ_1 =0,θ_2 =π g(φ)=sin φ,∣g′(φ)∣=∣cos φ∣ ∫_(−π) ^π δ(sin φ)dφ=Σ_k (1/(∣g′(θ_k )))=(1/1)+(1/1)=2 ∫_(−π) ^π (1/1)δ(sin φ)dφ=(1/2)×2=1](https://www.tinkutara.com/question/Q222806.png)
$${J}_{\mathrm{0}} \left(\mathrm{2}{x}\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} {e}^{{i}\mathrm{2}{x}\:\mathrm{sin}\:\phi} \mathrm{d}\phi \\ $$$$\int_{−\infty} ^{\infty} {J}_{\mathrm{0}} \left(\mathrm{2}{x}\right){dx}=\int_{−\infty} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} {e}^{{i}\mathrm{2}{x}\:\mathrm{sin}\:\phi} \mathrm{d}\phi\mathrm{d}{x} \\ $$$$=\int_{−\pi} ^{\pi} \int_{−\infty} ^{\infty} {e}^{{i}\mathrm{2}{x}\:\mathrm{sin}\:\phi} \mathrm{d}\phi\mathrm{d}{x} \\ $$$$\int_{−\infty} ^{\infty} {e}^{{ikx}} {dx}=\mathrm{2}\pi{k}\delta\left({k}\right),{k}=\mathrm{2}\:\mathrm{sin}\:\phi \\ $$$$\int_{−\infty} ^{\infty} {e}^{{i}\mathrm{2}{x}\:\mathrm{sin}\:\phi} {dx}=\mathrm{2}\pi\delta\left(\mathrm{2}\:\mathrm{sin}\:\phi\right) \\ $$$$\delta\left(\mathrm{2}\:\mathrm{sin}\:\phi\right)=\frac{\mathrm{1}}{\mathrm{2}}\delta\left(\mathrm{sin}\:\phi\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} \mathrm{2}\pi\centerdot\frac{\mathrm{1}}{\mathrm{2}}\delta\left(\mathrm{sin}\:\phi\right)\mathrm{d}\phi=\int_{−\pi} ^{\pi} \frac{\mathrm{1}}{\mathrm{2}}\delta\left(\mathrm{sin}\:\phi\right)\mathrm{d}\phi \\ $$$$\phi\in\left[−\pi,\pi\right],\mathrm{sin}\:\phi=\mathrm{0}\:\mathrm{at}\:\phi=\mathrm{0},\pm\pi \\ $$$$\phi=−\pi\equiv\pi\left(\mathrm{periodicity}\right) \\ $$$$\mathrm{zeros}:\theta_{\mathrm{1}} =\mathrm{0},\theta_{\mathrm{2}} =\pi \\ $$$${g}\left(\phi\right)=\mathrm{sin}\:\phi,\mid{g}'\left(\phi\right)\mid=\mid\mathrm{cos}\:\phi\mid \\ $$$$\int_{−\pi} ^{\pi} \delta\left(\mathrm{sin}\:\phi\right)\mathrm{d}\phi=\underset{{k}} {\sum}\frac{\mathrm{1}}{\mid{g}'\left(\theta_{{k}} \right)}=\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{2} \\ $$$$\int_{−\pi} ^{\pi} \frac{\mathrm{1}}{\mathrm{1}}\delta\left(\mathrm{sin}\:\phi\right)\mathrm{d}\phi=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}=\mathrm{1} \\ $$
Answered by TonyLin last updated on 08/Jul/25
$$\int_{−\infty} ^{\infty} {J}_{\mathrm{0}} \left(\mathrm{2}{x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {J}_{\mathrm{0}} \left(\mathrm{2}{x}\right){dx}\:\:\left({even}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {J}_{\mathrm{0}} \left(\mathrm{2}{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {J}_{\mathrm{0}} \left({x}\right){dx} \\ $$$$\Rightarrow\int_{−\infty} ^{\infty} {J}_{\mathrm{0}} \left(\mathrm{2}{x}\right){dx}=\int_{\mathrm{0}} ^{\infty} {J}_{\mathrm{0}} \left({x}\right){dx} \\ $$$$\mathrm{Given}\:\mathrm{that}\: \\ $$$$\mathscr{L}\left\{{J}_{\mathrm{0}} \left({x}\right)\right\}=\frac{\mathrm{1}}{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}=\int_{\mathrm{0}} ^{\infty} {J}_{\mathrm{0}} \left({x}\right){e}^{−{sx}} {dx} \\ $$$$\mathrm{L}{et}\:{s}=\mathrm{0} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} {J}_{\mathrm{0}} \left({x}\right){dx}=\mathrm{1} \\ $$$$\Rightarrow\int_{−\infty} ^{\infty} {J}_{\mathrm{0}} \left(\mathrm{2}{x}\right){dx}=\mathrm{1} \\ $$