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Question-199719

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Question Number 199719 by sonukgindia last updated on 08/Nov/23
Answered by witcher3 last updated on 08/Nov/23
I_a =I_b I_a =∫_0 ^π e^(sin(x)) cos(cos(x))dx+∫_π ^(2π) e^(sin(x)) (cos(cos(x))dx =∫_0 ^π e^(sin(x)) (cos(cos(x))+e^(−sin(x)) (cos(cos(x))dx I_b =∫_0 ^π e^(cos(x)) (cos(sin(x))+e^(cos(x)) (cos(sin(x))dx =∫_0 ^(π/2) (e^(cos(x)) (cos(sin(x))+e^(cos(x)) (cos(sin(x))dx′x→(π/2)−x +∫_(π/2) ^π (........)..x→(π/2)+x =2∫_0 ^(π/2) e^(sin(x)) cos(cos(x))dx+2∫_0 ^(π/2) e^(−sin(x)) (cos(cos(x))dx =∫_0 ^π e^(sin(x)) cos(cos(x))+e^(−sin(x)) cos(cos(x))dx=I_b I_a =2∫_0 ^π (cos(cos(x)ch(sin(x))dx =4∫_0 ^(π/2) (cos(cos(x))ch(sin(x))dx =4∫_0 ^1 cos(cos(x))Σ_(n≥0) ((sin^(2n) (x))/((2n)!))dx =4∫_0 ^(π/2) cos(cos(x))dx+4Σ_(n≥1) ∫_0 ^(π/2) cos(cos(x))Σ_(n≥1) ((sin^(2n) (x))/((2n)!))dx 4A+4B;A=2πJ_0 (1).easy B=Σ_(n≥1) ∫_0 ^1 cos(t)((√(1−t^2 )))^(2n−1) dt =Σ_(n≥1) ∫_0 ^1 cos(t)(1−t^2 )^(n−(1/2)) dt ∫_0 ^1 cos(t)(1−t^2 )^a dt=E =Σ_(n≥0) (((−1)^n )/((2n)!))∫_0 ^1 t^(2n) (1−t^2 )^a dt y=t^2 ⇔Σ_(n≥0) (((−1)^n )/((2n)!))(1/2)∫_0 ^1 y^(n−(1/2)) (1−y)^a dy =Σ_(n≥0) (((−1)^n )/(2(2n)!))((Γ(n+(1/2))Γ(1+a))/(Γ(n+a+(3/2)))) J_m (z)Σ_(n≥0) (((−1)^n z^(2n+m) )/(2^(2n+m) (n)!(n+m)!));bassel function =((Γ(1+a))/2)Σ_(n≥0) (((−1)^n Γ(n+(1/2)))/((n+a+(1/2))!.(2n)!))... (2n)!=2^n .n!.2^n Π_(k=0) ^(n−1) (k+(1/2)) =2^(2n) .n!.((Γ(n+(1/2)))/(Γ((1/2)))) E=((Γ(1+a)2^(a+(1/2)) )/2).(√π)Σ_(n≥0) (((−1)^n )/(2^(2n+a+(1/2)) .n!.(n+a+(1/2)))) =2^(a−(1/2)) (√π)Γ(1+a)J_(a+(1/2)) (1) B=Σ_(n≥1) 2^(n−1) (√π)Γ(1+(n/2))J_n (1) I_a =2πJ_0 (1)+Σ_(n≥1) 2^(n−1) (√π)Γ(1+(n/2))J_n (1)
$$\mathrm{I}_{\mathrm{a}} =\mathrm{I}_{\mathrm{b}} \\ $$$$\mathrm{I}_{\mathrm{a}} =\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{\mathrm{sin}\left(\mathrm{x}\right)} \mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx}+\int_{\pi} ^{\mathrm{2}\pi} \mathrm{e}^{\mathrm{sin}\left(\mathrm{x}\right)} \left(\mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx}\right. \\ $$$$=\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{\mathrm{sin}\left(\mathrm{x}\right)} \left(\mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)+\mathrm{e}^{−\mathrm{sin}\left(\mathrm{x}\right)} \left(\mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx}\right.\right. \\ $$$$\mathrm{I}_{\mathrm{b}} =\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{\mathrm{cos}\left(\mathrm{x}\right)} \left(\mathrm{cos}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)+\mathrm{e}^{\mathrm{cos}\left(\mathrm{x}\right)} \left(\mathrm{cos}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)\mathrm{dx}\right.\right. \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{e}^{\mathrm{cos}\left(\mathrm{x}\right)} \left(\mathrm{cos}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)+\mathrm{e}^{\mathrm{cos}\left(\mathrm{x}\right)} \left(\mathrm{cos}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)\mathrm{dx}'\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}−\mathrm{x}\right.\right.\right. \\ $$$$+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \left(……..\right)..\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}+\mathrm{x} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{e}^{\mathrm{sin}\left(\mathrm{x}\right)} \mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{e}^{−\mathrm{sin}\left(\mathrm{x}\right)} \left(\mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx}\right. \\ $$$$=\int_{\mathrm{0}} ^{\pi} \mathrm{e}^{\mathrm{sin}\left(\mathrm{x}\right)} \mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)+\mathrm{e}^{−\mathrm{sin}\left(\mathrm{x}\right)} \mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx}=\mathrm{I}_{\mathrm{b}} \\ $$$$\mathrm{I}_{\mathrm{a}} =\mathrm{2}\int_{\mathrm{0}} ^{\pi} \left(\mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\mathrm{ch}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)\mathrm{dx}\right.\right. \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{ch}\left(\mathrm{sin}\left(\mathrm{x}\right)\right)\mathrm{dx}\right. \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\left(\mathrm{2n}\right)!}\mathrm{dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\mathrm{dx}+\mathrm{4}\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{sin}^{\mathrm{2n}} \left(\mathrm{x}\right)}{\left(\mathrm{2n}\right)!}\mathrm{dx} \\ $$$$\mathrm{4A}+\mathrm{4B};\mathrm{A}=\mathrm{2}\pi\mathrm{J}_{\mathrm{0}} \left(\mathrm{1}\right).\mathrm{easy} \\ $$$$\mathrm{B}=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{t}\right)\left(\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\right)^{\mathrm{2n}−\mathrm{1}} \mathrm{dt} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{t}\right)\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{t}\right)\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{a}} \mathrm{dt}=\mathrm{E} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{\mathrm{2n}} \left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{a}} \mathrm{dt} \\ $$$$\mathrm{y}=\mathrm{t}^{\mathrm{2}} \Leftrightarrow\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{y}^{\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{y}\right)^{\mathrm{a}} \mathrm{dy} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2}\left(\mathrm{2n}\right)!}\frac{\Gamma\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}+\mathrm{a}\right)}{\Gamma\left(\mathrm{n}+\mathrm{a}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$\mathrm{J}_{\mathrm{m}} \left(\mathrm{z}\right)\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{z}^{\mathrm{2n}+\mathrm{m}} }{\mathrm{2}^{\mathrm{2n}+\mathrm{m}} \left(\mathrm{n}\right)!\left(\mathrm{n}+\mathrm{m}\right)!};\mathrm{bassel}\:\mathrm{function} \\ $$$$=\frac{\Gamma\left(\mathrm{1}+\mathrm{a}\right)}{\mathrm{2}}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \Gamma\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\mathrm{n}+\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}\right)!.\left(\mathrm{2n}\right)!}… \\ $$$$\left(\mathrm{2n}\right)!=\mathrm{2}^{\mathrm{n}} .\mathrm{n}!.\mathrm{2}^{\mathrm{n}} \underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{k}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}^{\mathrm{2n}} .\mathrm{n}!.\frac{\Gamma\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\mathrm{E}=\frac{\Gamma\left(\mathrm{1}+\mathrm{a}\right)\mathrm{2}^{\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}}.\sqrt{\pi}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2}^{\mathrm{2n}+\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}} .\mathrm{n}!.\left(\mathrm{n}+\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\mathrm{2}^{\mathrm{a}−\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\pi}\Gamma\left(\mathrm{1}+\mathrm{a}\right)\mathrm{J}_{\mathrm{a}+\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}\right) \\ $$$$\mathrm{B}=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{2}^{\mathrm{n}−\mathrm{1}} \sqrt{\pi}\Gamma\left(\mathrm{1}+\frac{\mathrm{n}}{\mathrm{2}}\right)\mathrm{J}_{\mathrm{n}} \left(\mathrm{1}\right) \\ $$$$\mathrm{I}_{\mathrm{a}} =\mathrm{2}\pi\mathrm{J}_{\mathrm{0}} \left(\mathrm{1}\right)+\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\mathrm{2}^{\mathrm{n}−\mathrm{1}} \sqrt{\pi}\Gamma\left(\mathrm{1}+\frac{\mathrm{n}}{\mathrm{2}}\right)\mathrm{J}_{\mathrm{n}} \left(\mathrm{1}\right) \\ $$$$ \\ $$

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