Question Number 36429 by prof Abdo imad last updated on 02/Jun/18
$${let}\:\:\varphi\left(\lambda\right)\:=\:\int_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left({x}\right){dx}\:{with}\:\lambda>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:\varphi\left(\lambda\right) \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:\varphi^{'} \left(\lambda\right). \\ $$
Commented byabdo mathsup 649 cc last updated on 03/Jun/18
![let integrate by parts u^′ =1+(1/x^2 ) and v^′ = arctan(x) ϕ(λ) = [(1−(1/x))arctanx]_(λ/π) ^(π/λ) −∫_(λ/π) ^(π/λ) (1−(1/x)) (dx/(1+x^2 )) =(1−(λ/π))arctan((π/λ)) −(1−(π/λ)) arctan((λ/π)) −∫_(λ/π) ^(π/λ) (((x−1)dx)/(x(1+x^2 ))) but ∫_(λ/π) ^(π/λ) ((x−1)/(x(1+x^2 )))dx = ∫_(λ/π) ^(π/λ) (dx/(1+x^2 )) −∫_(λ/π) ^(π/λ) (dx/(x(1+x^2 ))) = arctan((π/λ)) −arctan((λ/π)) −∫_(λ/π) ^(π/λ) (dx/(x(1+x^2 ))) but F(x)= (1/(x(1+x^2 ))) = (a/x) +((bx+c)/(x^2 +1)) a= 1 and lim_(x→+∞) x F(x)=0= a+b ⇒b=−1 c=0 ⇒ ∫_(λ/π) ^(π/λ) (dx/(x(1+x^2 ))) = ∫_(λ/π) ^(π/λ) { (1/x) −(x/(1+x^2 ))}dx =[ln ∣x∣ −(1/2)ln(1+x^2 )]_(λ/π) ^(π/λ) =[ln∣ (x/(√(1+x^2 )))∣]_(λ/π) ^(π/λ) = ln∣ (π/(λ(√(1+((π/λ))^2 ))))∣−ln∣ (λ/(π(√(1+((λ/π))^2 )))) ϕ(λ) =(1−(λ/π))arctan((π/λ)) −(1−(π/λ))arctan((λ/π)) −ln∣(π/(λ(√(1+((π/λ))^2 ))))∣ +ln∣ (λ/(π(√(1+((λ/π))^2 ))))∣ .](Q36523.png)
$${let}\:{integrate}\:{by}\:{parts}\:\:{u}^{'} \:=\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:{and} \\ $$ $${v}^{'} \:=\:{arctan}\left({x}\right)\: \\ $$ $$\varphi\left(\lambda\right)\:=\:\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){arctanx}\right]_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \: \\ $$ $$−\int_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \:\:\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$ $$=\left(\mathrm{1}−\frac{\lambda}{\pi}\right){arctan}\left(\frac{\pi}{\lambda}\right)\:−\left(\mathrm{1}−\frac{\pi}{\lambda}\right)\:{arctan}\left(\frac{\lambda}{\pi}\right) \\ $$ $$−\int_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \:\:\:\frac{\left({x}−\mathrm{1}\right){dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{but} \\ $$ $$\int_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \:\:\:\frac{{x}−\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:=\:\int_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:−\int_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$ $$=\:{arctan}\left(\frac{\pi}{\lambda}\right)\:−{arctan}\left(\frac{\lambda}{\pi}\right)\:−\int_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \:\:\:\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{but} \\ $$ $${F}\left({x}\right)=\:\frac{\mathrm{1}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\:=\:\frac{{a}}{{x}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$ $${a}=\:\mathrm{1}\:\:\:\:{and}\:{lim}_{{x}\rightarrow+\infty} {x}\:{F}\left({x}\right)=\mathrm{0}=\:{a}+{b}\:\Rightarrow{b}=−\mathrm{1} \\ $$ $${c}=\mathrm{0}\:\Rightarrow\:\int_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \:\:\frac{{dx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:=\:\int_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \:\:\left\{\:\frac{\mathrm{1}}{{x}}\:−\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right\}{dx} \\ $$ $$=\left[{ln}\:\mid{x}\mid\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \\ $$ $$=\left[{ln}\mid\:\:\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\mid\right]_{\frac{\lambda}{\pi}} ^{\frac{\pi}{\lambda}} \:=\:{ln}\mid\:\frac{\pi}{\lambda\sqrt{\mathrm{1}+\left(\frac{\pi}{\lambda}\right)^{\mathrm{2}} }}\mid−{ln}\mid\:\frac{\lambda}{\pi\sqrt{\mathrm{1}+\left(\frac{\lambda}{\pi}\right)^{\mathrm{2}} }} \\ $$ $$\varphi\left(\lambda\right)\:=\left(\mathrm{1}−\frac{\lambda}{\pi}\right){arctan}\left(\frac{\pi}{\lambda}\right)\:−\left(\mathrm{1}−\frac{\pi}{\lambda}\right){arctan}\left(\frac{\lambda}{\pi}\right) \\ $$ $$−{ln}\mid\frac{\pi}{\lambda\sqrt{\mathrm{1}+\left(\frac{\pi}{\lambda}\right)^{\mathrm{2}} }}\mid\:+{ln}\mid\:\:\frac{\lambda}{\pi\sqrt{\mathrm{1}+\left(\frac{\lambda}{\pi}\right)^{\mathrm{2}} }}\mid\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
$$\int_{\frac{\lambda}{\Pi}} ^{\frac{\Pi}{\lambda}} \:\left(\frac{\mathrm{1}+{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right){tan}^{−\mathrm{1}} {x}\:{dx} \\ $$ $${I}=\int\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){tan}^{−\mathrm{1}} {xdx} \\ $$ $$={tan}^{−\mathrm{1}} {x}\left({x}+\frac{−\mathrm{1}}{{x}}\right)−\int\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }×\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\right){dx} \\ $$ $$=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} {x}−\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$ $$=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} {x}\:−\int\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$ $$=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} {x}−\int\frac{{dx}}{{x}}+\mathrm{2}\int\frac{{xdx}}{{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$ $$=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} {x}−{lnx}+\int\frac{{dt}}{{t}\left({t}+\mathrm{1}\right)} \\ $$ $$=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} {x}−{lnx}+\int\frac{{t}+\mathrm{1}−{t}}{\left({t}+\mathrm{1}\right){t}}{dt} \\ $$ $$=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} {x}−{lnx}+{ln}\left(\frac{{t}}{{t}+\mathrm{1}}\right) \\ $$ $$=\left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} {x}−{lnx}+{ln}\left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$ $${now}\:{put}\:{upper}\:{and}\:{lower}\:{limit} \\ $$