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Question Number 38718 by maxmathsup by imad last updated on 28/Jun/18
1) find f(x)=∫_0 ^π ln(2+x cosθ)dθ 2) calculate ∫_0 ^π ln(2 +cosθ)dθ
$$\left.\mathrm{1}\right)\:{find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} \:{ln}\left(\mathrm{2}+{x}\:{cos}\theta\right){d}\theta \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\pi} \:{ln}\left(\mathrm{2}\:\:+{cos}\theta\right){d}\theta \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 30/Jun/18
1) we have f^′ (x)= ∫_0 ^π ((cosθ)/(2+x cosθ))dθ changement tan((θ/2))=t ⇒f^′ (x)=∫_0 ^∞ (((1−t^2 )/(1+t^2 ))/(2+x((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) f^′ (x) = 2∫_0 ^∞ ((1−t^2 )/((1+t^2 )(2+2t^2 +x−xt^2 )))dt =2 ∫_0 ^∞ ((1−t^2 )/((1+t^2 ){(2−x)t^2 +x+2}))dt let f^′ (x) =∫_(−∞) ^(+∞) ((1−t^2 )/((1+t^2 ){ (2−x)t^(2 ) +x+2})) dt let ϕ(z)= ((1−z^2 )/((z^2 +1){ (2−x)z^2 +x+2})) if x≠2 ϕ(z) = ((1−z^2 )/((2−x)(z^2 +1){ z^2 +((2+x)/(2−x))})) case 1 ((2+x)/(2−x))>0 ⇒ ϕ(z)= ((1−z^2 )/((2−x)(z−i)(z+i)(z−i(√((2+x)/(2−x))))(z+i(√((2+x)/(2−x)))))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,i) +Res(ϕ,i(√((2+x)/(2−x))))} Res(ϕ,i) = (2/((2−x)(2i)(−1 +((2+x)/(2−x))))) = ((−i(2−x))/((2−x)(−2+x+2+x))) =((−i)/(2x)) Res(ϕ,i(√((2+x)/(2−x)))) = ((1 +((2+x)/(2−x)))/((2−x)(−((2+x)/(2−x)) +1)2i(√((2+x)/(2−x))))) = (4/((2−x)^2 (((−2−x+2−x)/(2−x)))2i(√((2+x)/(2−x))))) = ((−2i)/((2−x)(−2x)((√(2+x))/( (√(2−x)))))) = (i/( (√(4−x^2 )))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ ((−i)/(2x)) + (i/( (√(4−x^2 ))))} =(π/x) +((2π)/( (√(4−x^2 )))) =f^′ (x) ⇒ f(x) =πln∣x∣ + 2π ∫ (dx/( (√(4−x^2 )))) +c changement x=2u give ∫ (dx/( (√(4−x^2 )))) = ∫ ((2du)/(2(√(1−u^2 )))) = arcsin((x/2)) ⇒f(x)=π ln∣x∣ +2π arcsin((x/2)) +c f(1) =2π (π/2) +c = π^2 +c ⇒c=f(1) −π^2 ⇒ f(x) =π ln∣x∣ +2π arcsin((x/2)) +f(1)−π^2 .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{cos}\theta}{\mathrm{2}+{x}\:{cos}\theta}{d}\theta\:{changement} \\ $$$${tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}\:\Rightarrow{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}+{x}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${f}^{'} \left({x}\right)\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} \:+{x}−{xt}^{\mathrm{2}} \right)}{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left\{\left(\mathrm{2}−{x}\right){t}^{\mathrm{2}} \:+{x}+\mathrm{2}\right\}}{dt}\:{let}\: \\ $$$${f}^{'} \left({x}\right)\:=\int_{−\infty} ^{+\infty} \:\:\:\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left\{\:\left(\mathrm{2}−{x}\right){t}^{\mathrm{2}\:} \:+{x}+\mathrm{2}\right\}}\:{dt}\:\:\:{let} \\ $$$$\varphi\left({z}\right)=\:\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left\{\:\left(\mathrm{2}−{x}\right){z}^{\mathrm{2}} \:+{x}+\mathrm{2}\right\}}\:{if}\:{x}\neq\mathrm{2} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\left(\mathrm{2}−{x}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left\{\:{z}^{\mathrm{2}} \:+\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}\right\}} \\ $$$${case}\:\mathrm{1}\:\:\:\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}>\mathrm{0}\:\Rightarrow \\ $$$$\varphi\left({z}\right)=\:\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\left(\mathrm{2}−{x}\right)\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{i}\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}\right)\left({z}+{i}\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{i}\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}\right)\right\} \\ $$$${Res}\left(\varphi,{i}\right)\:=\:\frac{\mathrm{2}}{\left(\mathrm{2}−{x}\right)\left(\mathrm{2}{i}\right)\left(−\mathrm{1}\:+\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}\right)} \\ $$$$=\:\frac{−{i}\left(\mathrm{2}−{x}\right)}{\left(\mathrm{2}−{x}\right)\left(−\mathrm{2}+{x}+\mathrm{2}+{x}\right)}\:=\frac{−{i}}{\mathrm{2}{x}} \\ $$$${Res}\left(\varphi,{i}\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}\right)\:=\:\:\frac{\mathrm{1}\:+\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}{\left(\mathrm{2}−{x}\right)\left(−\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}\:+\mathrm{1}\right)\mathrm{2}{i}\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}} \\ $$$$=\:\:\frac{\mathrm{4}}{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} \left(\frac{−\mathrm{2}−{x}+\mathrm{2}−{x}}{\mathrm{2}−{x}}\right)\mathrm{2}{i}\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}} \\ $$$$=\:\frac{−\mathrm{2}{i}}{\left(\mathrm{2}−{x}\right)\left(−\mathrm{2}{x}\right)\frac{\sqrt{\mathrm{2}+{x}}}{\:\sqrt{\mathrm{2}−{x}}}}\:=\:\frac{{i}}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{−{i}}{\mathrm{2}{x}}\:+\:\frac{{i}}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\right\} \\ $$$$=\frac{\pi}{{x}}\:+\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:={f}^{'} \left({x}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\pi{ln}\mid{x}\mid\:\:+\:\mathrm{2}\pi\:\int\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:\:+{c}\:\:{changement} \\ $$$${x}=\mathrm{2}{u}\:{give}\:\int\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:=\:\int\:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }} \\ $$$$=\:{arcsin}\left(\frac{{x}}{\mathrm{2}}\right)\:\Rightarrow{f}\left({x}\right)=\pi\:{ln}\mid{x}\mid\:+\mathrm{2}\pi\:{arcsin}\left(\frac{{x}}{\mathrm{2}}\right)\:+{c} \\ $$$${f}\left(\mathrm{1}\right)\:=\mathrm{2}\pi\:\frac{\pi}{\mathrm{2}}\:+{c}\:=\:\pi^{\mathrm{2}} \:+{c}\:\Rightarrow{c}={f}\left(\mathrm{1}\right)\:−\pi^{\mathrm{2}} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=\pi\:{ln}\mid{x}\mid\:+\mathrm{2}\pi\:{arcsin}\left(\frac{{x}}{\mathrm{2}}\right)\:+{f}\left(\mathrm{1}\right)−\pi^{\mathrm{2}} \:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 30/Jun/18
case 2 ((2+x)/(2−x)) <0 we follow the same method..
$${case}\:\mathrm{2}\:\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}\:<\mathrm{0}\:\:{we}\:{follow}\:{the}\:{same}\:{method}.. \\ $$
Commented by math khazana by abdo last updated on 30/Jun/18
error from line 15 Res(ϕ,i(√((2+x)/(2−x)))) = (i/(x(√(4−x^2 )))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ ((−i)/(2x)) +(i/(x(√(4−x^2 ))))} = (π/x) −((2π)/(x(√(4−x^2 )))) =f^′ (x) ⇒ f(x) =πln∣x∣ −2π ∫ (dx/(x(√(4−x^2 )))) .changement x= 2sint ⇒ ∫ (dx/(x(√(4−x^2 )))) = ∫ ((2cost)/(2sint 2 cost))dt =(1/2) ∫ (dt/(sint )) then chang.tan((t/2))=u =(1/2) ∫ (1/((2u)/(1+u^2 ))) ((2du)/(1+u^2 )) =(1/2) ∫ (du/u) =(1/2)ln∣tan((t/2))) =(1/2)ln∣ tan((1/2) arcsin((x/2)))∣⇒ f(x)=πln∣x∣ −πln∣ tan((1/2)arcsin((x/2))∣ +c f(1) = c ⇒ f(x)=πln∣x∣ −πln∣tan((1/2)arcsin((x/2))∣ + ∫_0 ^π ln(2+cosθ)dθ
$${error}\:{from}\:{line}\:\mathrm{15} \\ $$$${Res}\left(\varphi,{i}\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}\right)\:=\:\:\frac{{i}}{{x}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\frac{−{i}}{\mathrm{2}{x}}\:\:+\frac{{i}}{{x}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\right\} \\ $$$$=\:\frac{\pi}{{x}}\:\:−\frac{\mathrm{2}\pi}{{x}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:={f}^{'} \left({x}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\pi{ln}\mid{x}\mid\:\:−\mathrm{2}\pi\:\int\:\:\:\:\:\frac{{dx}}{{x}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:\:.{changement} \\ $$$${x}=\:\mathrm{2}{sint}\:\Rightarrow \\ $$$$\int\:\:\:\frac{{dx}}{{x}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:=\:\int\:\:\:\frac{\mathrm{2}{cost}}{\mathrm{2}{sint}\:\mathrm{2}\:{cost}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{{dt}}{{sint}\:}\:{then}\:{chang}.{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u} \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{du}}{{u}}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{tan}\left(\frac{{t}}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\:{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{arcsin}\left(\frac{{x}}{\mathrm{2}}\right)\right)\mid\Rightarrow \\ $$$${f}\left({x}\right)=\pi{ln}\mid{x}\mid\:−\pi{ln}\mid\:{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}{arcsin}\left(\frac{{x}}{\mathrm{2}}\right)\mid\:\:+{c}\right. \\ $$$${f}\left(\mathrm{1}\right)\:=\:{c}\:\Rightarrow \\ $$$${f}\left({x}\right)=\pi{ln}\mid{x}\mid\:−\pi{ln}\mid{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}{arcsin}\left(\frac{{x}}{\mathrm{2}}\right)\mid\:+\:\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{2}+{cos}\theta\right){d}\theta\right. \\ $$

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