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Question Number 37348 by math khazana by abdo last updated on 12/Jun/18
calculate ∫_0 ^(2π) (dt/(p +cost)) with p>1
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dt}}{{p}\:+{cost}}\:\:{with}\:{p}>\mathrm{1} \\ $$
Commented by prof Abdo imad last updated on 13/Jun/18
I = ∫_0 ^π (dt/(p+cost)) + ∫_π ^(2π) (dt/(p +cost)) =K +H changement tan((t/2))=x give K = ∫_0 ^∞ (1/(p +((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) =2 ∫_0 ^∞ (dx/(p+px^2 +1−x^2 )) = 2 ∫_0 ^∞ (dx/(p+1 +(p−1)x^2 )) =(2/(p+1))∫_0 ^∞ (dx/(1+((p−1)/(p+1))x^2 )) chang.(√((p−1)/(p+1))) x=u give K = (2/(p+1)) ∫_0 ^(+∞) (1/(1+u^2 )) (√((p+1)/(p−1)))du = (2/( (√(p^2 −1)))) .(π/2) = (π/( (√(p^2 −1)))) . changement t=π +x give H = ∫_0 ^π (dx/(p−cosx)) =_(tan((x/2))=t) ∫_0 ^∞ (1/(p−((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^∞ ((2dt)/(p +pt^2 −1+t^2 )) = _(x→0) =∫_0 ^∞ ((2dt)/(p−1 +(p+1)t^2 )) =(2/(p−1)) ∫_0 ^∞ (dt/(1+((p+1)/(p−1))t^2 )) =_((√((p+1)/(p−1))) t=u) (2/(p−1)) ∫_0 ^∞ (1/(1+u^2 )) (√((p−1)/(p+1))) du = (2/( (√(p^2 −1)))) .(π/2) = (π/( (√(p^2 −1)))) ⇒ I = (π/( (√(p^2 −1)))) +(π/( (√(p^2 −1)))) I =((2π)/( (√(p^2 −1)))) with p>1 .
$${I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dt}}{{p}+{cost}}\:+\:\int_{\pi} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{p}\:+{cost}}\:={K}\:+{H} \\ $$$${changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}\:{give}\: \\ $$$${K}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{p}\:+\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{p}+{px}^{\mathrm{2}} \:+\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{{p}+\mathrm{1}\:+\left({p}−\mathrm{1}\right){x}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{{p}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\mathrm{1}+\frac{{p}−\mathrm{1}}{{p}+\mathrm{1}}{x}^{\mathrm{2}} } \\ $$$${chang}.\sqrt{\frac{{p}−\mathrm{1}}{{p}+\mathrm{1}}}\:{x}={u}\:{give} \\ $$$${K}\:=\:\frac{\mathrm{2}}{{p}+\mathrm{1}}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\sqrt{\frac{{p}+\mathrm{1}}{{p}−\mathrm{1}}}{du} \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}}\:.\frac{\pi}{\mathrm{2}}\:=\:\frac{\pi}{\:\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}}\:. \\ $$$${changement}\:{t}=\pi\:+{x}\:{give} \\ $$$${H}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\frac{{dx}}{{p}−{cosx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{{p}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dt}}{{p}\:+{pt}^{\mathrm{2}} \:−\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dt}}{{p}−\mathrm{1}\:+\left({p}+\mathrm{1}\right){t}^{\mathrm{2}} }\:\:=\frac{\mathrm{2}}{{p}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+\frac{{p}+\mathrm{1}}{{p}−\mathrm{1}}{t}^{\mathrm{2}} } \\ $$$$=_{\sqrt{\frac{{p}+\mathrm{1}}{{p}−\mathrm{1}}}\:{t}={u}} \:\:\:\:\frac{\mathrm{2}}{{p}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\sqrt{\frac{{p}−\mathrm{1}}{{p}+\mathrm{1}}}\:{du} \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}}\:.\frac{\pi}{\mathrm{2}}\:\:=\:\frac{\pi}{\:\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow \\ $$$${I}\:=\:\:\frac{\pi}{\:\sqrt{{p}^{\mathrm{2}} \:−\mathrm{1}}}\:+\frac{\pi}{\:\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}}\: \\ $$$${I}\:=\frac{\mathrm{2}\pi}{\:\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}}\:\:\:{with}\:{p}>\mathrm{1}\:. \\ $$
Commented by prof Abdo imad last updated on 13/Jun/18
Residus method changement e^(it) =z give I =∫_(∣z∣=1) (1/(p +((z+z^(−1) )/2))) (dz/(iz)) = ∫_(∣z∣=1) ((−2idz)/(z(2p +z+z^(−1) ))) = ∫_(∣z∣=1) ((−2idz)/(2pz +z^2 +1)) =∫_(∣z∣=1) ((−2idz)/(z^2 +2pz +1)) let ϕ(z) = ((−2i)/(z^2 +2pz +1)) poles of ϕ? Δ^′ =p^2 −1 ⇒ z_1 =−p +(√(p^2 −1)) z_2 =−p−(√(p^2 −1)) ∣z_1 ∣−1 =p−(√(p^2 −1)) −1 =p−1 −(√(p^2 −1)) (p−1)^2 −(p^2 −1) =p^2 −2p +1 −p^2 +1 =−2p+2 =−2(p−1)<0 ⇒∣z_1 ∣<1 ∣z_2 ∣−1 =p +(√(p^2 −1))−1>0 (to eliminate from Residus) ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ ,z_1 ) Res(ϕ,z_1 ) =((−2i)/(z_1 −z_2 )) = ((−2i)/(2(√(p^2 −1)))) = ((−i)/( (√(p^2 −1)))) ∫_(∣z∣=1) ϕ(z)dz =2iπ(((−i)/( (√(p^2 −1))))) = ((2π)/( (√(p^2 −1)))) ⇒ ★ I = ((2π)/( (√(p^2 −1)))) ★
$${Residus}\:{method} \\ $$$${changement}\:{e}^{{it}} ={z}\:{give} \\ $$$${I}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{{p}\:+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{{z}\left(\mathrm{2}{p}\:+{z}+{z}^{−\mathrm{1}} \right)}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{\mathrm{2}{pz}\:+{z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{{z}^{\mathrm{2}} \:+\mathrm{2}{pz}\:+\mathrm{1}}\:\:{let}\:\varphi\left({z}\right)\:=\:\frac{−\mathrm{2}{i}}{{z}^{\mathrm{2}} \:+\mathrm{2}{pz}\:+\mathrm{1}} \\ $$$${poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} \:={p}^{\mathrm{2}} \:−\mathrm{1}\:\Rightarrow\:{z}_{\mathrm{1}} =−{p}\:+\sqrt{{p}^{\mathrm{2}} −\mathrm{1}} \\ $$$${z}_{\mathrm{2}} =−{p}−\sqrt{{p}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:={p}−\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}\:−\mathrm{1}\:={p}−\mathrm{1}\:−\sqrt{{p}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\left({p}−\mathrm{1}\right)^{\mathrm{2}} \:−\left({p}^{\mathrm{2}} −\mathrm{1}\right)\:={p}^{\mathrm{2}} \:−\mathrm{2}{p}\:+\mathrm{1}\:−{p}^{\mathrm{2}} \:+\mathrm{1}\: \\ $$$$=−\mathrm{2}{p}+\mathrm{2}\:=−\mathrm{2}\left({p}−\mathrm{1}\right)<\mathrm{0}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:={p}\:+\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}>\mathrm{0}\:\left({to}\:{eliminate}\:{from}\right. \\ $$$$\left.{Residus}\right) \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi\:,{z}_{\mathrm{1}} \right)\:\:\: \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\frac{−\mathrm{2}{i}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }\:=\:\frac{−\mathrm{2}{i}}{\mathrm{2}\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}}\:=\:\frac{−{i}}{\:\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\frac{−{i}}{\:\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}}\right) \\ $$$$=\:\frac{\mathrm{2}\pi}{\:\sqrt{{p}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow \\ $$$$\bigstar\:\:{I}\:\:=\:\:\frac{\mathrm{2}\pi}{\:\sqrt{{p}^{\mathrm{2}} \:−\mathrm{1}}}\:\bigstar \\ $$

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