Question Number 38458 by maxmathsup by imad last updated on 25/Jun/18
$${let}\:\mid{x}\mid<\mathrm{1}\:\:{calculate}\:\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right){dt} \\ $$ $$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \right){dt} \\ $$ $$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{sin}\theta\:{t}^{\mathrm{2}} \right){dt}\:. \\ $$
Commented bymath khazana by abdo last updated on 28/Jun/18
![we have F^′ (x)= ∫_0 ^1 (t^2 /(1+xt^2 ))dt =(1/x) ∫_0 ^1 ((xt^2 +1 −1)/(xt^2 +1))dt =(1/x) −(1/x) ∫_0 ^1 (dt/(1+xt^2 )) so if 0<x<1changement (√x)t=u give ∫_0 ^1 (dt/(1+xt^2 )) = ∫_0 ^(√x) (1/(1+u^2 )) (du/(√x)) =(1/(√x)) arctan((√x)) ⇒ F^′ (x)= (1/x) −((arctan((√x)))/(x(√x))) ⇒ F(x) = ∫_1 ^x (dt/t) − ∫_1 ^x ((arctan((√t)))/(t(√t)))dt +c =ln(x) −∫_1 ^x ((arctan((√t)))/(t(√t)))dt +c but chang.(√t)=u give ∫_1 ^x ((arctan((√t)))/(t(√t)))dt =∫_1 ^(√x) ((arctan(u))/(u^2 .u)) 2u du =2 ∫_1 ^(√x) ((arctan(u))/u^2 ) du by parts ∫_1 ^(√x) ((arctanu)/u^2 ) du = [−(1/u) arctan(u)]_1 ^(√x) +∫_1 ^(√x) (1/(u(1+u^2 )))du = (π/4) −((arctan((√x)))/(√x)) + ∫_1 ^(√x) ( (1/u) −(u/(1+u^2 )))du =(π/4) −((arctan((√x)))/(√x)) +[ln((u/(√(1+u^2 ))))]_1 ^(√x) =(π/4) −((arctan((√x)))/(√x)) +ln(((√x)/(√(1+x)))) +ln((√2)) ⇒ F(x)= ln(x) −(π/2) +((2arctan((√x)))/(√x)) −2ln(((√x)/(√(1+x)))) −ln(2) +c F(1)=∫_0 ^1 ln(1+t^2 )dt =−(π/2) +(π/2) −2ln((1/(√2))) −ln(2) +c=c ⇒ F(x)= −(π/2) +((2 arctan((√x)))/(√x)) +ln(1+x)−ln(2) + ∫_0 ^1 ln(1+t^2 )dt by parts ∫_0 ^1 ln(1+t^2 )dt =[tln(1+t^2 )]_0 ^1 −∫_0 ^1 ((t 2t)/(1+t^2 ))dt =ln(2) −2 ∫_0 ^1 ((1+t^2 −1)/(1+t^2 ))dt =ln(2) −2 +2 ∫_0 ^1 (dt/(1+t^2 )) =ln(2) −2 +(π/2) ⇒ F(x)= ((2 arctan((√x)))/(√x)) +ln(1+x) −2 with 0<x<1](Q38660.png)
$${we}\:{have}\:{F}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{xt}^{\mathrm{2}} }{dt} \\ $$ $$=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xt}^{\mathrm{2}} \:+\mathrm{1}\:−\mathrm{1}}{{xt}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$ $$=\frac{\mathrm{1}}{{x}}\:\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{xt}^{\mathrm{2}} }\:\:\:{so}\:{if}\:\mathrm{0}<{x}<\mathrm{1}{changement}\:\sqrt{{x}}{t}={u}\:{give} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+{xt}^{\mathrm{2}} }\:\:=\:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:\frac{{du}}{\sqrt{{x}}} \\ $$ $$=\frac{\mathrm{1}}{\sqrt{{x}}}\:{arctan}\left(\sqrt{{x}}\right)\:\Rightarrow \\ $$ $${F}^{'} \left({x}\right)=\:\frac{\mathrm{1}}{{x}}\:−\frac{{arctan}\left(\sqrt{{x}}\right)}{{x}\sqrt{{x}}}\:\Rightarrow \\ $$ $${F}\left({x}\right)\:=\:\int_{\mathrm{1}} ^{{x}} \:\:\frac{{dt}}{{t}}\:\:−\:\int_{\mathrm{1}} ^{{x}} \:\:\:\frac{{arctan}\left(\sqrt{{t}}\right)}{{t}\sqrt{{t}}}{dt}\:+{c} \\ $$ $$={ln}\left({x}\right)\:−\int_{\mathrm{1}} ^{{x}} \:\:\:\frac{{arctan}\left(\sqrt{{t}}\right)}{{t}\sqrt{{t}}}{dt}\:+{c}\:{but}\:{chang}.\sqrt{{t}}={u} \\ $$ $${give}\:\int_{\mathrm{1}} ^{{x}} \:\:\frac{{arctan}\left(\sqrt{{t}}\right)}{{t}\sqrt{{t}}}{dt}\:=\int_{\mathrm{1}} ^{\sqrt{{x}}} \:\:\:\:\frac{{arctan}\left({u}\right)}{{u}^{\mathrm{2}} .{u}}\:\mathrm{2}{u}\:{du} \\ $$ $$=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{{x}}} \:\:\frac{{arctan}\left({u}\right)}{{u}^{\mathrm{2}} }\:{du}\:\:\:{by}\:{parts} \\ $$ $$\int_{\mathrm{1}} ^{\sqrt{{x}}} \:\:\:\frac{{arctanu}}{{u}^{\mathrm{2}} }\:{du}\:=\:\:\left[−\frac{\mathrm{1}}{{u}}\:{arctan}\left({u}\right)\right]_{\mathrm{1}} ^{\sqrt{{x}}} \\ $$ $$+\int_{\mathrm{1}} ^{\sqrt{{x}}} \:\:\:\frac{\mathrm{1}}{{u}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du}\:=\:\frac{\pi}{\mathrm{4}}\:−\frac{{arctan}\left(\sqrt{{x}}\right)}{\sqrt{{x}}} \\ $$ $$+\:\int_{\mathrm{1}} ^{\sqrt{{x}}} \left(\:\frac{\mathrm{1}}{{u}}\:−\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\right){du} \\ $$ $$=\frac{\pi}{\mathrm{4}}\:−\frac{{arctan}\left(\sqrt{{x}}\right)}{\sqrt{{x}}}\:\:+\left[{ln}\left(\frac{{u}}{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\right)\right]_{\mathrm{1}} ^{\sqrt{{x}}} \\ $$ $$=\frac{\pi}{\mathrm{4}}\:−\frac{{arctan}\left(\sqrt{{x}}\right)}{\sqrt{{x}}}\:+{ln}\left(\frac{\sqrt{{x}}}{\sqrt{\mathrm{1}+{x}}}\right)\:+{ln}\left(\sqrt{\mathrm{2}}\right)\:\Rightarrow \\ $$ $${F}\left({x}\right)=\:{ln}\left({x}\right)\:−\frac{\pi}{\mathrm{2}}\:+\frac{\mathrm{2}{arctan}\left(\sqrt{{x}}\right)}{\sqrt{{x}}}\:−\mathrm{2}{ln}\left(\frac{\sqrt{{x}}}{\sqrt{\mathrm{1}+{x}}}\right) \\ $$ $$−{ln}\left(\mathrm{2}\right)\:+{c} \\ $$ $${F}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt}\:=−\frac{\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}\:−\mathrm{2}{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right) \\ $$ $$−{ln}\left(\mathrm{2}\right)\:+{c}={c}\:\Rightarrow \\ $$ $${F}\left({x}\right)=\:−\frac{\pi}{\mathrm{2}}\:\:+\frac{\mathrm{2}\:{arctan}\left(\sqrt{{x}}\right)}{\sqrt{{x}}}\:\:+{ln}\left(\mathrm{1}+{x}\right)−{ln}\left(\mathrm{2}\right) \\ $$ $$+\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt}\:\:{by}\:{parts} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt}\:=\left[{tln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}\:\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$ $$={ln}\left(\mathrm{2}\right)\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{t}^{\mathrm{2}} \:−\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$ $$={ln}\left(\mathrm{2}\right)\:−\mathrm{2}\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:={ln}\left(\mathrm{2}\right)\:−\mathrm{2}\:\:+\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$ $${F}\left({x}\right)=\:\frac{\mathrm{2}\:{arctan}\left(\sqrt{{x}}\right)}{\sqrt{{x}}}\:+{ln}\left(\mathrm{1}+{x}\right)\:−\mathrm{2}\:{with}\:\mathrm{0}<{x}<\mathrm{1} \\ $$ $$ \\ $$
Commented bymath khazana by abdo last updated on 28/Jun/18
![if −1<x<0 we have F^′ (x)=(1/x) −(1/x) ∫_0 ^1 (dt/(1+xt^2 )) ∫_0 ^1 (dt/(1+xt^2 )) =∫_0 ^1 (dt/(1−(−x)t^2 )) =_((√(−x))t=u) ∫_0 ^(√(−x)) (1/(1−u^2 )) (du/(√(−x))) =(1/(√(−x))) ∫_0 ^(√(−x)) (du/(1−u^2 )) =(1/(2(√(−x)))) ∫_0 ^(√(−x)) ( (1/(1+u)) +(1/(1−u)))du = (1/(2(√(−x)))) [ln(((1+u)/(1−u)))]_0 ^(√(−x)) = (1/(2(√(−x)))) ln(((1+(√(−x)))/(1−(√(−x)))))⇒ F^′ (x) = (1/x) −(1/(2x(√(−x))))ln(((1+(√(−x)))/(1−(√(−x))))) ⇒ F(x)= ln(−x) − ∫_1 ^x (1/(2t(√(−t)))) ln(((1+(√(−t)))/(1−(√(−t)))))dt +c ...be continued...](Q38664.png)
$${if}\:−\mathrm{1}<{x}<\mathrm{0}\:{we}\:{have}\:{F}^{'} \left({x}\right)=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+{xt}^{\mathrm{2}} } \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{xt}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}−\left(−{x}\right){t}^{\mathrm{2}} }\:=_{\sqrt{−{x}}{t}={u}} \:\int_{\mathrm{0}} ^{\sqrt{−{x}}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}−{u}^{\mathrm{2}} }\:\frac{{du}}{\sqrt{−{x}}} \\ $$ $$=\frac{\mathrm{1}}{\sqrt{−{x}}}\:\int_{\mathrm{0}} ^{\sqrt{−{x}}} \:\:\:\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{x}}}\:\int_{\mathrm{0}} ^{\sqrt{−{x}}} \left(\:\frac{\mathrm{1}}{\mathrm{1}+{u}}\:+\frac{\mathrm{1}}{\mathrm{1}−{u}}\right){du} \\ $$ $$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{x}}}\:\left[{ln}\left(\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\right)\right]_{\mathrm{0}} ^{\sqrt{−{x}}} \:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{x}}}\:{ln}\left(\frac{\mathrm{1}+\sqrt{−{x}}}{\mathrm{1}−\sqrt{−{x}}}\right)\Rightarrow \\ $$ $${F}^{'} \left({x}\right)\:=\:\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{−{x}}}{ln}\left(\frac{\mathrm{1}+\sqrt{−{x}}}{\mathrm{1}−\sqrt{−{x}}}\right)\:\Rightarrow \\ $$ $${F}\left({x}\right)=\:{ln}\left(−{x}\right)\:\:−\:\int_{\mathrm{1}} ^{{x}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{t}\sqrt{−{t}}}\:{ln}\left(\frac{\mathrm{1}+\sqrt{−{t}}}{\mathrm{1}−\sqrt{−{t}}}\right){dt}\:+{c} \\ $$ $$...{be}\:{continued}... \\ $$
Commented bymath khazana by abdo last updated on 28/Jun/18
$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \right){dt}\:={F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{2}{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)}{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}}\:+{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$ $$−\mathrm{2}\:=\mathrm{2}\sqrt{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\:+{ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{2}\right)\:−\mathrm{2}\:. \\ $$