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Question Number 146780 by tabata last updated on 15/Jul/21

$$findbyresidue{\int }_0^{2\pi }\frac{d\theta }{1+ksin\theta },0<k<1$$

Answered by mathmax by abdo last updated on 15/Jul/21

$$\mathrm{\Phi }={\int }_0^{2\pi }\frac{\mathrm{dx}}{1+\mathrm{ksinx}}\Rightarrow \mathrm{\Phi }{=}_{{\mathrm{e}}^{\mathrm{ix}}=\mathrm{z}}{\int }_{\mid \mathrm{z}\mid =1}\frac{\mathrm{dz}}{\mathrm{iz}(1+\mathrm{k}\frac{\mathrm{z}-{\mathrm{z}}^{-1}}{2\mathrm{i}})}$$ $$={\int }_{\mid \mathrm{z}\mid =1}\frac{2\mathrm{idz}}{\mathrm{iz}(2\mathrm{i}+\mathrm{kz}-{\mathrm{kz}}^{-1})}={\int }_{\mid \mathrm{z}\mid =1}\frac{2\mathrm{dz}}{2\mathrm{iz}+{\mathrm{kz}}^2-\mathrm{k}}$$ $$\phi \left(\mathrm{z}\right)=\frac{2}{{\mathrm{kz}}^2+2\mathrm{iz}-\mathrm{k}}$$ $${\mathrm{\Delta }}^{{}^{\prime }}=-1+{\mathrm{k}}^2<0\Rightarrow {\mathrm{z}}_1=\frac{-\mathrm{i}+\mathrm{i}\sqrt{1-{\mathrm{k}}^2}}{\mathrm{k}}\mathrm{and}{\mathrm{z}}_2=\frac{-\mathrm{i}-\mathrm{i}\sqrt{1-{\mathrm{k}}^2}}{\mathrm{k}}$$ $$\mid {\mathrm{z}}_1\mid -1=\frac{\sqrt{1+1-{\mathrm{k}}^2}}{\mathrm{k}}-1=\frac{\sqrt{2-{\mathrm{k}}^2}-\mathrm{k}}{\mathrm{k}}>0\Rightarrow \mid {\mathrm{z}}_1\mid >1\Rightarrow \mathrm{Res}=0$$ $$\mid {\mathrm{z}}_2\mid -1=\frac{\sqrt{1+1-{\mathrm{k}}^2}}{\mathrm{k}}-1>1\Rightarrow \mathrm{Res}=0\Rightarrow {\int }_{\mathrm{R}}\phi \left(2\right)\mathrm{dz}=0\Rightarrow \mathrm{\Phi }=0$$

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