Question Number 65133 by turbo msup by abdo last updated on 25/Jul/19
$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sint}\:+{xcost}\right){dt} \\ $$$${x}\:{real}. \\ $$
Commented by mathmax by abdo last updated on 27/Jul/19
![we have f^′ (x) =∫_0 ^(π/4) ((cost)/(sint +xcost))dt changement tan((t/2)) =u give f^′ (x) = ∫_0 ^((√2)−1) (((1−u^2 )/(1+u^2 ))/(((2u)/(1+u^2 )) +x((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) =2∫_0 ^((√2)−1) ((1−u^2 )/((1+u^2 )(2u +x−xu^2 )))du =2 ∫_0 ^((√2)−1) ((u^2 −1)/((u^2 +1)(xu^2 −2u −x)))du let decompose F(u) =((u^2 −1)/((u^2 +1)(xu^2 −2u −x))) xu^2 −2u−x =0→Δ^′ =1+x^2 >0 ⇒u_1 (x)=((1+(√(1+x^2 )))/x) u_2 (x)=((1−(√(1+x^2 )))/x) ⇒F(u) =((u^2 −1)/(x(u^2 +1)(u−u_1 )(u−u_2 ))) F(u) =(a/(u−u_1 )) +(b/(u−u_2 )) +((cu +d)/(u^2 +1)) a =lim_(u→u_1 ) (u−u_(1 ) )F(u) =((u_1 ^2 −1)/(x(u_1 ^2 +1)(u_1 −u_2 ))) b =lim_(u→u_2 ) (u−u_2 )F(u) =((u_2 ^2 −1)/(x(u_2 ^2 +1)(u_2 −u_1 ))) lim_(u→+∞) uF(u) =0 =a+b +c ⇒c =−a−b F(0) =(1/x) =−(a/u_1 ) −(b/u_2 ) +d ⇒d =(1/x) +(a/u_1 ) +(b/u_2 ) ⇒ F(u) =(a/(u−u_1 )) +(b/(u−u_2 )) +((−(a+b)u +(1/x)+(a/u_1 )+(b/u_2 ))/(u^2 +1)) ⇒ f^′ (x) =2 ∫_0 ^((√2)−1) ((a/(u−u_1 )) +(b/(u−u_2 )) +((−(a+b)u +λ_x )/(u^(2 ) +1)) )du =2a [ln∣u−u_1 ∣]_0 ^((√2)−1) +2b [ln∣u−u_2 ∣]_0 ^((√2)−1) −(a+b)[ln(u^2 +1)]_0 ^((√2)−1) +2λ_x [arctanu]_0 ^((√2)−1) =2a ln∣(((√2)−1−u_1 )/u_1 )∣ +2b ln∣(((√2)−1−u_2 )/u_2 )∣−(a+b)ln( ((√2)−1)^2 +1) +2λ_x (π/8) ⇒ f(x) =2a ∫ ln∣(((√2)−1−u_1 (x))/(u_1 (x)))∣dx +2b∫ ln∣(((√2)−1−u_2 (x))/(u_2 (x)))∣dx −(a+b)ln(4−2(√2))x +(π/4) ∫ λ_x dx +c ....be continued....](https://www.tinkutara.com/question/Q65283.png)
$${we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{cost}}{{sint}\:+{xcost}}{dt}\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={u}\:{give} \\ $$$${f}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+{x}\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\mathrm{2}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{2}{u}\:+{x}−{xu}^{\mathrm{2}} \right)}{du} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({xu}^{\mathrm{2}} −\mathrm{2}{u}\:−{x}\right)}{du}\:{let}\:{decompose} \\ $$$${F}\left({u}\right)\:=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({xu}^{\mathrm{2}} −\mathrm{2}{u}\:−{x}\right)} \\ $$$${xu}^{\mathrm{2}} −\mathrm{2}{u}−{x}\:=\mathrm{0}\rightarrow\Delta^{'} \:=\mathrm{1}+{x}^{\mathrm{2}} \:>\mathrm{0}\:\Rightarrow{u}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}} \\ $$$${u}_{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\:\Rightarrow{F}\left({u}\right)\:=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{{x}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)} \\ $$$$ \\ $$$$ \\ $$$${F}\left({u}\right)\:=\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{u}\rightarrow{u}_{\mathrm{1}} } \:\:\left({u}−{u}_{\mathrm{1}\:} \right){F}\left({u}\right)\:=\frac{{u}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{1}}{{x}\left({u}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}\right)\left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right)} \\ $$$${b}\:={lim}_{{u}\rightarrow{u}_{\mathrm{2}} } \:\:\:\:\left({u}−{u}_{\mathrm{2}} \right){F}\left({u}\right)\:=\frac{{u}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{1}}{{x}\left({u}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}_{\mathrm{2}} −{u}_{\mathrm{1}} \right)} \\ $$$${lim}_{{u}\rightarrow+\infty} {uF}\left({u}\right)\:=\mathrm{0}\:={a}+{b}\:+{c}\:\Rightarrow{c}\:=−{a}−{b} \\ $$$${F}\left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{{x}}\:=−\frac{{a}}{{u}_{\mathrm{1}} }\:−\frac{{b}}{{u}_{\mathrm{2}} }\:+{d}\:\Rightarrow{d}\:=\frac{\mathrm{1}}{{x}}\:+\frac{{a}}{{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}_{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:+\frac{−\left({a}+{b}\right){u}\:+\frac{\mathrm{1}}{{x}}+\frac{{a}}{{u}_{\mathrm{1}} }+\frac{{b}}{{u}_{\mathrm{2}} }}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\left(\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:+\frac{−\left({a}+{b}\right){u}\:+\lambda_{{x}} }{{u}^{\mathrm{2}\:} +\mathrm{1}}\:\right){du} \\ $$$$=\mathrm{2}{a}\:\left[{ln}\mid{u}−{u}_{\mathrm{1}} \mid\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:+\mathrm{2}{b}\:\left[{ln}\mid{u}−{u}_{\mathrm{2}} \mid\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:−\left({a}+{b}\right)\left[{ln}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$+\mathrm{2}\lambda_{{x}} \:\:\:\left[{arctanu}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$=\mathrm{2}{a}\:{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−{u}_{\mathrm{1}} }{{u}_{\mathrm{1}} }\mid\:+\mathrm{2}{b}\:{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−{u}_{\mathrm{2}} }{{u}_{\mathrm{2}} }\mid−\left({a}+{b}\right){ln}\left(\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right) \\ $$$$+\mathrm{2}\lambda_{{x}} \frac{\pi}{\mathrm{8}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\mathrm{2}{a}\:\int\:\:{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−{u}_{\mathrm{1}} \left({x}\right)}{{u}_{\mathrm{1}} \left({x}\right)}\mid{dx}\:+\mathrm{2}{b}\int\:{ln}\mid\frac{\sqrt{\mathrm{2}}−\mathrm{1}−{u}_{\mathrm{2}} \left({x}\right)}{{u}_{\mathrm{2}} \left({x}\right)}\mid{dx} \\ $$$$−\left({a}+{b}\right){ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right){x}\:+\frac{\pi}{\mathrm{4}}\:\int\:\:\lambda_{{x}} \:{dx}\:\:+{c}\:….{be}\:{continued}…. \\ $$$$ \\ $$