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Question Number 219872 by SdC355 last updated on 03/May/25
prove ∫ Y_(−(3/2)) (z) dz=((4sin(z)+((z𝚪((1/2),−iz))/( (√(−iz))))+((z𝚪((1/2),iz))/( (√(iz)))))/( (√(2πz))))+C
$$\mathrm{prove} \\ $$$$\int\:\:{Y}_{−\frac{\mathrm{3}}{\mathrm{2}}} \left({z}\right)\:\mathrm{d}{z}=\frac{\mathrm{4sin}\left({z}\right)+\frac{{z}\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}},−\boldsymbol{{i}}{z}\right)}{\:\sqrt{−\boldsymbol{{i}}{z}}}+\frac{{z}\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}},\boldsymbol{{i}}{z}\right)}{\:\sqrt{\boldsymbol{{i}}{z}}}}{\:\sqrt{\mathrm{2}\pi{z}}}+{C} \\ $$
Answered by MrGaster last updated on 03/May/25
Y_(−ν) =(−1)^ν Y_ν (z)⇒Y_(−(3/2)) (z)=(−1)^(3/2) Y_(3/2) =iY_(2/3) (z) Y_(3/2) (z)=−(√(2/(πz)))(((cos z)/z)+sin z)+(z^(3/2) /( (√(2π))))(Γ((1/2),iz)+Γ((1/2),−iz)) ∫Y_(−(3/2)) (z)dz=i∫Y_(3/2) (z)dz ∫Y_(−(3/2)) (z)dz=−(√(2/π))∫((cos z+sin z)/z^(1/2) )dz+(1/( (√(2π))))∫(Γ((1/2),iz)+Γ((1/2),−iz)) ∫((cos z)/z^(3/2) )dz=−2((sin z)/( (√z)))+2(√π)(((Γ((1/2),iz))/( (√(−iz))))+((Γ((1/2),−iz))/( (√(iz))))) ∫((sin z)/( (√z)))dz=(√(π/2))(Si(z)−Ci(z))⇒∫((sin z)/( (√z)))dz=(√(2/π))(((Γ((1/2),iz))/( (√(iz))))−((Γ((1/2),−iz))/( (√(−iz))))) ∫Γ((1/2),±iz)z^(1/2) dz=(2/3)z^(3/2) Γ((1/2),±iz)∓((2i)/3)z^(3/2) e^(∓iz) ⇒∫ Y_(−(3/2)) (z) dz=((4sin(z)+((z𝚪((1/2),−iz))/( (√(−iz))))+((z𝚪((1/2),iz))/( (√(iz)))))/( (√(2πz))))+C
$${Y}_{−\nu} =\left(−\mathrm{1}\right)^{\nu} {Y}_{\nu} \left({z}\right)\Rightarrow{Y}_{−\frac{\mathrm{3}}{\mathrm{2}}} \left({z}\right)=\left(−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {Y}_{\frac{\mathrm{3}}{\mathrm{2}}} ={iY}_{\frac{\mathrm{2}}{\mathrm{3}}} \left({z}\right) \\ $$$${Y}_{\frac{\mathrm{3}}{\mathrm{2}}} \left({z}\right)=−\sqrt{\frac{\mathrm{2}}{\pi{z}}}\left(\frac{\mathrm{cos}\:{z}}{{z}}+\mathrm{sin}\:{z}\right)+\frac{{z}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\:\sqrt{\mathrm{2}\pi}}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},{iz}\right)+\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},−{iz}\right)\right) \\ $$$$\int{Y}_{−\frac{\mathrm{3}}{\mathrm{2}}} \left({z}\right){dz}={i}\int{Y}_{\frac{\mathrm{3}}{\mathrm{2}}} \left({z}\right){dz} \\ $$$$\int{Y}_{−\frac{\mathrm{3}}{\mathrm{2}}} \left({z}\right){dz}=−\sqrt{\frac{\mathrm{2}}{\pi}}\int\frac{\mathrm{cos}\:{z}+\mathrm{sin}\:{z}}{{z}^{\frac{\mathrm{1}}{\mathrm{2}}} }{dz}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\int\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},{iz}\right)+\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},−{iz}\right)\right) \\ $$$$\int\frac{\mathrm{cos}\:{z}}{{z}^{\frac{\mathrm{3}}{\mathrm{2}}} }{dz}=−\mathrm{2}\frac{\mathrm{sin}\:{z}}{\:\sqrt{{z}}}+\mathrm{2}\sqrt{\pi}\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},{iz}\right)}{\:\sqrt{−{iz}}}+\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},−{iz}\right)}{\:\sqrt{{iz}}}\right) \\ $$$$\int\frac{\mathrm{sin}\:{z}}{\:\sqrt{{z}}}{dz}=\sqrt{\frac{\pi}{\mathrm{2}}}\left(\mathrm{Si}\left({z}\right)−\mathrm{Ci}\left({z}\right)\right)\Rightarrow\int\frac{\mathrm{sin}\:{z}}{\:\sqrt{{z}}}{dz}=\sqrt{\frac{\mathrm{2}}{\pi}}\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},{iz}\right)}{\:\sqrt{{iz}}}−\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},−{iz}\right)}{\:\sqrt{−{iz}}}\right) \\ $$$$\int\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},\pm{iz}\right){z}^{\frac{\mathrm{1}}{\mathrm{2}}} {dz}=\frac{\mathrm{2}}{\mathrm{3}}{z}^{\frac{\mathrm{3}}{\mathrm{2}}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},\pm{iz}\right)\mp\frac{\mathrm{2}{i}}{\mathrm{3}}{z}^{\frac{\mathrm{3}}{\mathrm{2}}} {e}^{\mp{iz}} \\ $$$$\Rightarrow\int\:\:{Y}_{−\frac{\mathrm{3}}{\mathrm{2}}} \left({z}\right)\:\mathrm{d}{z}=\frac{\mathrm{4sin}\left({z}\right)+\frac{{z}\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}},−\boldsymbol{{i}}{z}\right)}{\:\sqrt{−\boldsymbol{{i}}{z}}}+\frac{{z}\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}},\boldsymbol{{i}}{z}\right)}{\:\sqrt{\boldsymbol{{i}}{z}}}}{\:\sqrt{\mathrm{2}\pi{z}}}+{C} \\ $$
Commented by SdC355 last updated on 03/May/25
wow.............thx
$$\mathrm{wow}………….\mathrm{thx} \\ $$

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