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

find by residue ∫_0 ^( 2π) (dθ/(1+ksinθ)) ,0<k<1

$${find}\:{by}\:{residue}\:\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\frac{{d}\theta}{\mathrm{1}+{ksin}\theta}\:\:\:,\mathrm{0}<{k}<\mathrm{1} \\ $$

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

Φ=∫_0 ^(2π) (dx/(1+ksinx)) ⇒Φ=_(e^(ix) =z) ∫_(∣z∣=1) (dz/(iz(1+k((z−z^(−1) )/(2i))))) =∫_(∣z∣=1) ((2idz)/(iz(2i+kz−kz^(−1) ))) =∫_(∣z∣=1) ((2dz)/(2iz+kz^2 −k)) ϕ(z)=(2/(kz^2 +2iz−k)) Δ^′ =−1+k^2 <0 ⇒z_1 =((−i+i(√(1−k^2 )))/k) and z_2 =((−i−i(√(1−k^2 )))/k) ∣z_1 ∣−1=((√(1+1−k^2 ))/k)−1 =(((√(2−k^2 ))−k)/k)>0⇒∣z_1 ∣>1 ⇒Res=0 ∣z_2 ∣−1=((√(1+1−k^2 ))/k)−1>1 ⇒Res=0 ⇒∫_R ϕ(2)dz=0 ⇒Φ=0

$$\Phi=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{ksinx}}\:\Rightarrow\Phi=_{\mathrm{e}^{\mathrm{ix}} \:=\mathrm{z}} \:\:\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{dz}}{\mathrm{iz}\left(\mathrm{1}+\mathrm{k}\frac{\mathrm{z}−\mathrm{z}^{−\mathrm{1}} }{\mathrm{2i}}\right)} \\ $$ $$=\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2idz}}{\mathrm{iz}\left(\mathrm{2i}+\mathrm{kz}−\mathrm{kz}^{−\mathrm{1}} \right)}\:=\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{2dz}}{\mathrm{2iz}+\mathrm{kz}^{\mathrm{2}} −\mathrm{k}} \\ $$ $$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{2}}{\mathrm{kz}^{\mathrm{2}} \:+\mathrm{2iz}−\mathrm{k}} \\ $$ $$\Delta^{'} \:=−\mathrm{1}+\mathrm{k}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{i}+\mathrm{i}\sqrt{\mathrm{1}−\mathrm{k}^{\mathrm{2}} }}{\mathrm{k}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\frac{−\mathrm{i}−\mathrm{i}\sqrt{\mathrm{1}−\mathrm{k}^{\mathrm{2}} }}{\mathrm{k}} \\ $$ $$\mid\mathrm{z}_{\mathrm{1}} \mid−\mathrm{1}=\frac{\sqrt{\mathrm{1}+\mathrm{1}−\mathrm{k}^{\mathrm{2}} }}{\mathrm{k}}−\mathrm{1}\:=\frac{\sqrt{\mathrm{2}−\mathrm{k}^{\mathrm{2}} }−\mathrm{k}}{\mathrm{k}}>\mathrm{0}\Rightarrow\mid\mathrm{z}_{\mathrm{1}} \mid>\mathrm{1}\:\Rightarrow\mathrm{Res}=\mathrm{0} \\ $$ $$\mid\mathrm{z}_{\mathrm{2}} \mid−\mathrm{1}=\frac{\sqrt{\mathrm{1}+\mathrm{1}−\mathrm{k}^{\mathrm{2}} }}{\mathrm{k}}−\mathrm{1}>\mathrm{1}\:\Rightarrow\mathrm{Res}=\mathrm{0}\:\Rightarrow\int_{\mathrm{R}} \varphi\left(\mathrm{2}\right)\mathrm{dz}=\mathrm{0}\:\Rightarrow\Phi=\mathrm{0} \\ $$

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