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Question Number 59247 by maxmathsup by imad last updated on 06/May/19

let f(x) =∫_0 ^(π/2) ln(1−xcost)dt with ∣x∣<1 1) developp f at integr serie 2) find a explicit form of f(x) 3) find the values of integrals ∫_0 ^(π/2) ln(1−cost)dt and ∫_0 ^(π/2) ln(1+cost)dt 4) calculate U_n =∫_0 ^(π/2) ln(1−(2/n)cost)dt with n integr and n≥2 study the convergence of U_n and Σ U_n

$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{xcost}\right){dt}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$ $$\left.\mathrm{1}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$ $$\left.\mathrm{2}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{3}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{cost}\right){dt}\:\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}+{cost}\right){dt} \\ $$ $$\left.\mathrm{4}\right)\:{calculate}\:\:{U}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−\frac{\mathrm{2}}{{n}}{cost}\right){dt}\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{2} \\ $$ $${study}\:{the}\:{convergence}\:{of}\:{U}_{{n}} \:\:\:\:{and}\:\Sigma\:{U}_{{n}} \\ $$

Commented byMr X pcx last updated on 09/May/19

2) we have f^′ (x) = ∫_0 ^(π/2) ((−cost)/(1−xcost)) dt for x≠0 f^′ (x) =(1/x)∫_0 ^(π/2) ((1−xcost −1)/(1−xcost))dt =(π/(2x)) −(1/x) ∫_0 ^(π/2) (dt/(1−x cost)) but chang.tan((t/2)) =u give ∫_0 ^(π/2) (dt/(1−xcost)) =∫_0 ^1 (1/(1−x((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) =∫_0 ^1 ((2du)/(1+u^2 −x+xu^2 )) =∫_0 ^1 ((2du)/((1+x)u^2 +1−x)) =(2/(1+x)) ∫_0 ^1 (du/(u^2 +((1−x)/(1+x)))) =_(u=(√((1−x)/(1+x)))α) (2/(1+x)) ∫_0 ^(√((1+x)/(1−x))) (1/(((1−x)/(1+x))(1+α^2 )))(√((1−x)/(1+x)))dα =(2/(1−x)) ((√(1−x))/(√(1+x))) ∫_0 ^(√((1+x)/(1−x))) (dα/(1+α^2 )) =(2/(√(1−x^2 ))) [arctan(α)]_0 ^(√((1+x)/(1−x))) =(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/(x(√(1−x^2 )))) arctan((√((1+x)/(1−x))))dx +c ∫ (2/(x(√(1−x^2 )))) arctan((√((1+x)/(1−x))))dx =_(x =cost) ∫ (2/(cost.sint)) arctan((√((2cos^2 ((t/2)))/(2sin^2 ((t/2))))))(−sint)dt =−2 ∫ (1/(sint)) arctan((1/(tan((t/2)))))dt =−2 ∫ (1/(sint))((π/2) −(t/2))dt =−π ∫ (dt/(sint)) + ∫ (t/(sint)) dt ....be continued...

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{−{cost}}{\mathrm{1}−{xcost}}\:{dt} \\ $$ $${for}\:{x}\neq\mathrm{0}\:\:{f}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{1}−{xcost}\:−\mathrm{1}}{\mathrm{1}−{xcost}}{dt} \\ $$ $$=\frac{\pi}{\mathrm{2}{x}}\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{1}−{x}\:{cost}}\:\:{but} \\ $$ $${chang}.{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={u}\:{give} \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{1}−{xcost}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}−{x}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} −{x}+{xu}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{x}\right){u}^{\mathrm{2}} \:+\mathrm{1}−{x}} \\ $$ $$=\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}} \\ $$ $$=_{{u}=\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\alpha} \:\:\:\:\frac{\mathrm{2}}{\mathrm{1}+{x}}\:\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{d}\alpha \\ $$ $$=\frac{\mathrm{2}}{\mathrm{1}−{x}}\:\frac{\sqrt{\mathrm{1}−{x}}}{\sqrt{\mathrm{1}+{x}}}\:\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}} \:\:\:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} } \\ $$ $$=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\left[{arctan}\left(\alpha\right)\right]_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}} \\ $$ $$=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right)\:\Rightarrow \\ $$ $${f}^{'} \left({x}\right)\:=\frac{\pi}{\mathrm{2}{x}}\:−\frac{\mathrm{2}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right)\:\Rightarrow \\ $$ $${f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:−\int\:\:\:\frac{\mathrm{2}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right){dx}\:+{c} \\ $$ $$\int\:\:\frac{\mathrm{2}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right){dx} \\ $$ $$=_{{x}\:={cost}} \:\:\:\int\:\:\frac{\mathrm{2}}{{cost}.{sint}}\:{arctan}\left(\sqrt{\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{t}}{\mathrm{2}}\right)}}\right)\left(−{sint}\right){dt} \\ $$ $$=−\mathrm{2}\:\int\:\:\:\frac{\mathrm{1}}{{sint}}\:{arctan}\left(\frac{\mathrm{1}}{{tan}\left(\frac{{t}}{\mathrm{2}}\right)}\right){dt} \\ $$ $$=−\mathrm{2}\:\int\:\frac{\mathrm{1}}{{sint}}\left(\frac{\pi}{\mathrm{2}}\:−\frac{{t}}{\mathrm{2}}\right){dt} \\ $$ $$=−\pi\:\int\:\frac{{dt}}{{sint}}\:+\:\int\:\:\frac{{t}}{{sint}}\:{dt}\:....{be}\:{continued}... \\ $$ $$ \\ $$

Commented bymaxmathsup by imad last updated on 09/May/19

error at line 16 we have ∫ (2/(x(√(1−x^2 )))) arctan((√((1+x)/(1−x))))dx =−2 ∫ (1/(cost))((π/2)−(t/2))dt+c =−π ∫(dt/(cost)) + ∫ (t/(cost)) dt +c ⇒ f(x) =(π/2)ln∣x∣ +π∫ (dt/(cost)) −∫ (t/(cost)) dt +c ....

$${error}\:{at}\:{line}\:\mathrm{16}\:\:\:\:{we}\:{have}\:\int\:\:\:\:\frac{\mathrm{2}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right){dx} \\ $$ $$=−\mathrm{2}\:\int\:\frac{\mathrm{1}}{{cost}}\left(\frac{\pi}{\mathrm{2}}−\frac{{t}}{\mathrm{2}}\right){dt}+{c}\:=−\pi\:\int\frac{{dt}}{{cost}}\:+\:\int\:\:\frac{{t}}{{cost}}\:{dt}\:+{c}\:\Rightarrow \\ $$ $${f}\left({x}\right)\:=\frac{\pi}{\mathrm{2}}{ln}\mid{x}\mid\:+\pi\int\:\:\:\frac{{dt}}{{cost}}\:−\int\:\:\:\frac{{t}}{{cost}}\:{dt}\:\:+{c}\:.... \\ $$

Answered by Mr X pcx last updated on 09/May/19

we have ln^′ (1−u) =((−1)/(1−u)) =−Σ_(n=0) ^∞ u^n with ∣u∣<1 ⇒ln(1−u) =−Σ_(n=0) ^∞ (u^(n+1) /(n+1)) =−Σ_(n=1) ^∞ (u^n /n) we have ∣xcost∣<1 ⇒ ln(1−xcost) =−Σ_(n=1) ^∞ ((x^n cos^n t)/n) ⇒ ∫_0 ^(π/2) ln(1−xcost)dt =−Σ_(n=1) ^∞ (x^n /n) ∫_0 ^(π/2) cos^n (t) =−Σ_(n=1) ^∞ (A_n /n) x^n withA_n =∫_0 ^(π/2) cos^n t dt (integral of wallis) .

$${we}\:{have}\:{ln}^{'} \left(\mathrm{1}−{u}\right)\:=\frac{−\mathrm{1}}{\mathrm{1}−{u}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{u}^{{n}} \\ $$ $${with}\:\mid{u}\mid<\mathrm{1}\:\Rightarrow{ln}\left(\mathrm{1}−{u}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}} \\ $$ $$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{u}^{{n}} }{{n}}\:\:\:\:{we}\:{have}\:\mid{xcost}\mid<\mathrm{1}\:\Rightarrow \\ $$ $${ln}\left(\mathrm{1}−{xcost}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} \:{cos}^{{n}} {t}}{{n}}\:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{1}−{xcost}\right){dt}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{x}^{{n}} }{{n}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{{n}} \left({t}\right) \\ $$ $$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{A}_{{n}} }{{n}}\:{x}^{{n}} \:\:\:{withA}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{{n}} {t}\:{dt} \\ $$ $$\left({integral}\:{of}\:{wallis}\right)\:. \\ $$

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