Question Number 33884 by math khazana by abdo last updated on 26/Apr/18
$${let}\:{F}\left({x}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{arctan}\left({xtant}\right)}{{tant}}\:{dt}\:{find}\:{a}\:{simple} \\ $$$${form}\:{of}\:{f}\left({x}\right)\:. \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{arctan}\left(\mathrm{2}{tant}\right)}{{tant}}{dt}\:. \\ $$
Commented by abdo imad last updated on 28/Apr/18
![we have (dF/dx)(x)= ∫_0 ^(π/2) ((tant)/((1+x^2 tan^2 t)tant))dt = ∫_0 ^(π/2) (dt/(1+x^2 tan^2 t)) but changement tant =u give (dF/dx) =∫_0 ^∞ (1/((1+x^2 u^2 ))) (du/(1+u^2 )) let decompose F(u) = (1/((1+x^2 u^2 )(1+u^2 ))) = ((au+b)/(1+x^2 u^2 )) +((cu+d)/(1+u^2 )) F(−u)=F(u) ⇔((−au +b)/(1+x^2 u^2 )) +((−cu +d)/(1+u^2 )) =((au +b)/(1+x^2 u^2 )) +((cu +d)/(1+u^2 )) ⇒ a=0 and c=0 ⇒F(u)= (b/(1+x^2 u^2 )) +(d/(1+u^2 )) lim_(x→+∞) u^2 F(u)=0=(b/x^2 ) +d ⇒b+dx^2 =0 ⇒b=−dx^2 F(u) =((−dx^2 )/(1+x^2 u^2 )) + (d/(1+u^2 )) we have F(0)=1 =−dx^2 +d =(1−x^2 )d ⇒d=(1/(1−x^2 )) if x^2 ≠1 ⇒ F(u)=((−x^2 )/((1−x^2 )(1+x^2 u^2 ))) + (1/((1−x^2 )(1+u^2 ))) (dF/dx)(x)^ = ((−x^2 )/(1−x^2 )) ∫_0 ^∞ (du/(1+x^2 u^2 )) + (1/(1−x^2 )) ∫_0 ^∞ (du/(1+u^2 )) =_(xu =t) ((−x^2 )/(1−x^2 )) ∫_0 ^∞ (1/(1+t^2 )) (dt/x) +(1/(1−x^2 )) (π/2) =((−x)/(1−x^2 )) (π/2) +(π/2) (1/(1−x^2 )) =(π/(2(1−x^2 ))) (1−x) = (π/(2(1+x))) ⇒ F(x) = (π/2)∫_0 ^x ln(1+t)dt +λ but λ=F(0)=0 ⇒ F(x)=(π/2) ∫_0 ^x ln(1+t)dt =_(1+t=u) (π/2) ∫_1 ^(1+x) ln(u)du =(π/2)[uln(u)−u]_1 ^(1+x) =(π/2)((1+x)ln(1+x)−1−x +1) =(π/2)( (x+1)ln(x+1)−x) 2) ∫_0 ^(π/2) ((arctan(2tant))/(tant)) dt = F(2) =(π/2)(3ln(3) −2)](https://www.tinkutara.com/question/Q33965.png)
$${we}\:{have}\:\frac{{dF}}{{dx}}\left({x}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{tant}}{\left(\mathrm{1}+{x}^{\mathrm{2}} {tan}^{\mathrm{2}} {t}\right){tant}}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{x}^{\mathrm{2}} {tan}^{\mathrm{2}} {t}}\:\:\:\:{but}\:{changement}\:{tant}\:={u}\:{give} \\ $$$$\frac{{dF}}{{dx}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)}\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:{let}\:{decompose} \\ $$$${F}\left({u}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\:\frac{{au}+{b}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\:+\frac{{cu}+{d}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$${F}\left(−{u}\right)={F}\left({u}\right)\:\Leftrightarrow\frac{−{au}\:+{b}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\:+\frac{−{cu}\:+{d}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{{au}\:+{b}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\:+\frac{{cu}\:+{d}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\Rightarrow \\ $$$${a}=\mathrm{0}\:{and}\:{c}=\mathrm{0}\:\Rightarrow{F}\left({u}\right)=\:\frac{{b}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\:+\frac{{d}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow+\infty} {u}^{\mathrm{2}} {F}\left({u}\right)=\mathrm{0}=\frac{{b}}{{x}^{\mathrm{2}} }\:+{d}\:\Rightarrow{b}+{dx}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{b}=−{dx}^{\mathrm{2}} \\ $$$${F}\left({u}\right)\:=\frac{−{dx}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\:+\:\frac{{d}}{\mathrm{1}+{u}^{\mathrm{2}} }\:{we}\:{have}\:{F}\left(\mathrm{0}\right)=\mathrm{1}\:=−{dx}^{\mathrm{2}} \:+{d} \\ $$$$=\left(\mathrm{1}−{x}^{\mathrm{2}} \right){d}\:\Rightarrow{d}=\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:{if}\:{x}^{\mathrm{2}} \neq\mathrm{1}\:\Rightarrow \\ $$$${F}\left({u}\right)=\frac{−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} \right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$\frac{{dF}}{{dx}}\left({x}\right)^{} =\:\frac{−{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{x}^{\mathrm{2}} {u}^{\mathrm{2}} }\:\:+\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=_{{xu}\:={t}} \:\:\:\frac{−{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\frac{{dt}}{{x}}\:\:+\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\frac{\pi}{\mathrm{2}} \\ $$$$=\frac{−{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\frac{\pi}{\mathrm{2}}\:\:+\frac{\pi}{\mathrm{2}}\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:=\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}\:\left(\mathrm{1}−{x}\right)\:=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{x}\right)} \\ $$$$\Rightarrow\:{F}\left({x}\right)\:=\:\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} \:{ln}\left(\mathrm{1}+{t}\right){dt}\:\:+\lambda\:\:{but}\:\lambda={F}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{x}} {ln}\left(\mathrm{1}+{t}\right){dt}\:=_{\mathrm{1}+{t}={u}} \frac{\pi}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{1}+{x}} {ln}\left({u}\right){du} \\ $$$$=\frac{\pi}{\mathrm{2}}\left[{uln}\left({u}\right)−{u}\right]_{\mathrm{1}} ^{\mathrm{1}+{x}} \:=\frac{\pi}{\mathrm{2}}\left(\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)−\mathrm{1}−{x}\:+\mathrm{1}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\left(\:\left({x}+\mathrm{1}\right){ln}\left({x}+\mathrm{1}\right)−{x}\right) \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{arctan}\left(\mathrm{2}{tant}\right)}{{tant}}\:{dt}\:=\:{F}\left(\mathrm{2}\right)\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{3}{ln}\left(\mathrm{3}\right)\:−\mathrm{2}\right) \\ $$