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Question Number 149853 by 0731619 last updated on 07/Aug/21

Answered by mindispower last updated on 08/Aug/21

(1/π)∫_0 ^π e^(izcos(θ)) dθ=J_0 (z) bassel function cos^2 (θ)=((1+cos(2θ))/2) ⇔Im((1/2)∫_0 ^π e^(i(((1+cos(2x))/2))) dx.(1/2) =Im(e^(i/2) /2)(∫_0 ^(2π) e^(i((cos(x))/2)) dx) ∫_0 ^(2π) e^((icos(x))/2) dx=2∫_0 ^π e^(i((cos(x))/2)) dx we get (1/2)Ime^(i/2) ∫_0 ^π e^(i(1/2)cos(x)) dx using that ∫_0 ^π e^(i((cos(x))/2)) dx=πJ_0 ((1/2))∈R we get (1/2)Ime^(i/2) ∫_0 ^π e^(i((cos(x))/2)) dx=(π/2)sin((1/2))J_0 ((1/2))≃0.707

$$\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} {e}^{{izcos}\left(\theta\right)} {d}\theta={J}_{\mathrm{0}} \left({z}\right)\:\:{bassel}\:{function} \\ $$$${cos}^{\mathrm{2}} \left(\theta\right)=\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}} \\ $$$$\Leftrightarrow{Im}\left(\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {e}^{{i}\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)} {dx}.\frac{\mathrm{1}}{\mathrm{2}}\right. \\ $$$$={Im}\frac{{e}^{\frac{{i}}{\mathrm{2}}} }{\mathrm{2}}\left(\int_{\mathrm{0}} ^{\mathrm{2}\pi} {e}^{{i}\frac{{cos}\left({x}\right)}{\mathrm{2}}} {dx}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} {e}^{\frac{{icos}\left({x}\right)}{\mathrm{2}}} {dx}=\mathrm{2}\int_{\mathrm{0}} ^{\pi} {e}^{{i}\frac{{cos}\left({x}\right)}{\mathrm{2}}} {dx} \\ $$$${we}\:{get} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{Ime}^{\frac{{i}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\pi} {e}^{{i}\frac{\mathrm{1}}{\mathrm{2}}{cos}\left({x}\right)} {dx} \\ $$$${using}\:{that}\:\int_{\mathrm{0}} ^{\pi} {e}^{{i}\frac{{cos}\left({x}\right)}{\mathrm{2}}} {dx}=\pi{J}_{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\in\mathbb{R} \\ $$$${we}\:{get} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{Ime}^{\frac{{i}}{\mathrm{2}}} \int_{\mathrm{0}} ^{\pi} {e}^{{i}\frac{{cos}\left({x}\right)}{\mathrm{2}}} {dx}=\frac{\pi}{\mathrm{2}}{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){J}_{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\simeq\mathrm{0}.\mathrm{707} \\ $$$$ \\ $$

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