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Question Number 40621 by math khazana by abdo last updated on 25/Jul/18

let f(x)=∫_0 ^(π/2) ln(1+xcosθ)dθ 1) calculate f(1) 2) find a simple form of f(x) 3) developp f at ontehr serie

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\: \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{ontehr}\:{serie} \\ $$

Answered by math khazana by abdo last updated on 27/Jul/18

1) f(1)=∫_0 ^(π/2) ln(1+cosθ)dθ = I and let J = ∫_0 ^(π/2) ln(1−cosθ)dθ we have I +J = ∫_0 ^(π/2) ln(1−cos^2 θ)dθ = ∫_0 ^(π/2) ln(sin^2 θ)dθ =2 ∫_0 ^(π/2) ln(sinθ)dθ =2(−(π/2)ln(2))=−πln(2) I −J = ∫_0 ^(π/2) ln(((1+cosθ)/(1−cosθ)))dθ = ∫_0 ^(π/2) ln( ((2cos^2 ((θ/2)))/(2sin^2 ((θ/2)))))dθ = ∫_0 ^(π/2) −2 ln(tan((θ/2)))dθ =−2 ∫_0 ^(π/2) ln(tan((θ/2)))dθ =_((θ/2)=t) −4 ∫_0 ^(π/4) ln(tan(t))dt ∫_0 ^(π/4) ln(tant)dt =_(tant =u) ∫_0 ^1 ln(u) (du/(1+u^2 )) = ∫_0 ^1 ((ln(u))/(1+u^2 )) du =∫_0 ^1 ln(u) Σ_(n=0) ^∞ (−1)^n u^(2n) )du =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 u^(2n) ln(u)du =Σ_(n=0) ^∞ (−1)^n A_n A_n = [(1/(2n+1)) u^(2n+1) ln(u)]_0 ^1 −∫_0 ^1 (u^(2n) /((2n+1)))du =−(1/((2n+1)^2 )) ⇒ ∫_0 ^(π/4) ln(tant)dt = −Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) =−λ_0 (the value of λ_0 is known) ⇒ I −J =4 λ_0 ⇒ I +J = −π ln(2) and I −J =4λ_0 ⇒ 2I =−πln(2)+4λ_0 ⇒ I =−(π/2)ln(2) +2λ_0

$$\left.\mathrm{1}\right)\:{f}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta\:=\:{I}\:{and}\:{let}\: \\ $$$${J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{cos}\theta\right){d}\theta\:{we}\:{have}\: \\ $$$${I}\:+{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right){d}\theta\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({sin}\theta\right){d}\theta\:=\mathrm{2}\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right)=−\pi{ln}\left(\mathrm{2}\right) \\ $$$${I}\:−{J}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}+{cos}\theta}{\mathrm{1}−{cos}\theta}\right){d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left(\:\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}\right){d}\theta\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:−\mathrm{2}\:{ln}\left({tan}\left(\frac{\theta}{\mathrm{2}}\right)\right){d}\theta \\ $$$$=−\mathrm{2}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{ln}\left({tan}\left(\frac{\theta}{\mathrm{2}}\right)\right){d}\theta \\ $$$$=_{\frac{\theta}{\mathrm{2}}={t}} \:−\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({tan}\left({t}\right)\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({tant}\right){dt}\:=_{{tant}\:={u}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{ln}\left({u}\right)\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\left.=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left({u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }\:{du}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({u}\right)\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} \right){du} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{\mathrm{2}{n}} \:{ln}\left({u}\right){du}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{A}_{{n}} \\ $$$${A}_{{n}} =\:\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:{u}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({u}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{u}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}{du} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left({tant}\right){dt}\:=\:−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\lambda_{\mathrm{0}} \:\:\:\left({the}\:{value}\:{of}\:\lambda_{\mathrm{0}} \:{is}\:{known}\right)\:\Rightarrow \\ $$$${I}\:−{J}\:=\mathrm{4}\:\lambda_{\mathrm{0}} \:\:\Rightarrow\:{I}\:+{J}\:=\:−\pi\:{ln}\left(\mathrm{2}\right)\:{and} \\ $$$${I}\:−{J}\:=\mathrm{4}\lambda_{\mathrm{0}} \:\:\Rightarrow\:\mathrm{2}{I}\:=−\pi{ln}\left(\mathrm{2}\right)+\mathrm{4}\lambda_{\mathrm{0}} \:\Rightarrow \\ $$$${I}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+\mathrm{2}\lambda_{\mathrm{0}} \\ $$

Answered by math khazana by abdo last updated on 27/Jul/18

2) we have f^′ (x)= ∫_0 ^(π/2) ((cosθ)/(1+x cosθ)) =(1/x) ∫_0 ^(π/2) ((xcosθ +1 −1)/(1+xcosθ))dθ (x≠0) =(π/(2x)) −(1/x) ∫_0 ^(π/2) (dθ/(1+x cosθ)) but ∫_0 ^(π/2) (dθ/(1+xcosθ)) =_(tan((θ/2))=t) ∫_0 ^1 (1/(1+x((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^1 ((2dt)/(1+t^2 +x(1−t^2 ))) = ∫_0 ^1 ((2dt)/(1+x +(1−x)t^2 )) =(2/(1+x)) ∫_0 ^1 (dt/(1+((1−x)/(1+x))t^2 )) if ∣x∣<1 chang.(√((1−x)/(1+x)))t=u =(2/(1+x)) ∫_0 ^(√((1−x)/(1+x))) (1/(1+u^2 )) (√((1+x)/(1−x)))du = (2/(√(1−x^2 ))) arctan((√((1−x)/(1+x)))) ⇒ f^′ (x)= (π/(2x)) −(2/(x(√(1−x^2 )))) arctan((√((1−x)/(1+x))))⇒ f(x)=(π/2)ln∣x∣ −2 ∫ (1/(x(√(1−x^2 )))) arctan((√((1−x)/(1+x))))dx +c be continued....

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cos}\theta}{\mathrm{1}+{x}\:{cos}\theta} \\ $$$$=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{xcos}\theta\:+\mathrm{1}\:−\mathrm{1}}{\mathrm{1}+{xcos}\theta}{d}\theta\:\left({x}\neq\mathrm{0}\right)\: \\ $$$$=\frac{\pi}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\mathrm{1}+{x}\:{cos}\theta}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{\mathrm{1}+{xcos}\theta}\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{1}+{x}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+{x}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{x}\:+\left(\mathrm{1}−{x}\right){t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{t}^{\mathrm{2}} }\:\:{if}\:\mid{x}\mid<\mathrm{1}\:{chang}.\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{t}={u} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\:\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}{du} \\ $$$$=\:\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\:\frac{\pi}{\mathrm{2}{x}}\:\:−\frac{\mathrm{2}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\right)\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:−\mathrm{2}\:\int\:\:\:\:\frac{\mathrm{1}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\right){dx}\:+{c} \\ $$$${be}\:{continued}.... \\ $$

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