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let-pi-pi-1-1-x-2-arctan-x-dx-with-gt-0-1-find-a-simple-form-of-2-calculate-




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

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