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Question Number 38121 by maxmathsup by imad last updated on 22/Jun/18

let x>0 find F(x) = ∫_(−∞) ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt

$${let}\:{x}>\mathrm{0}\:{find}\:{F}\left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{arctan}\left({xt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\: \\ $$

Commented byprof Abdo imad last updated on 23/Jun/18

we have f^′ (x)= ∫_(−∞) ^(+∞) (t^2 /((1+t^2 )(1+x^2 t^4 )))dt let consider the complex function ϕ(z)= (z^2 /((z^2 +1)(x^2 z^4 +1))) .poles of ϕ? ϕ(z)=(z^2 /(x^2 (z^2 +1)(z^4 +(1/x^2 )))) = (z^2 /(x^2 (z−i)(z+i)(z^2 −(i/x))(z^2 +(i/x)))) = (z^2 /(x^2 (z−i)(z+i)(z −(1/(√x))e^((iπ)/4) )(z+(1/(√x))e^((iπ)/4) )(z−(1/(√x))e^(−((iπ)/4)) )(z+(1/(√x))e^(−((iπ)/4)) ))) so the poles of ϕ are +^− i, +^− (1/(√x))e^((iπ)/4) ,+^− (1/(√x))e^(−((iπ)/4)) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{ Res(ϕ,i) +Res(ϕ,(1/(√x))e^((iπ)/4) )+Res(ϕ,−(1/(√x))e^(−((iπ)/4)) )} Res(ϕ,i)= ((−1)/(2ix^2 (1 +(1/x^2 )))) = ((−1)/(2i(x^2 +1)))

$${we}\:{have}\:{f}^{'} \left({x}\right)=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} \right)}{dt} \\ $$ $${let}\:{consider}\:{the}\:{complex}\:{function} \\ $$ $$\varphi\left({z}\right)=\:\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} {z}^{\mathrm{4}} \:+\mathrm{1}\right)}\:.{poles}\:{of}\:\varphi? \\ $$ $$\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({z}^{\mathrm{4}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\: \\ $$ $$=\:\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}−{i}\right)\left({z}+{i}\right)\left({z}^{\mathrm{2}} \:−\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} \:+\frac{{i}}{{x}}\right)} \\ $$ $$=\:\frac{{z}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left({z}−{i}\right)\left({z}+{i}\right)\left({z}\:−\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$ $${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{i},\:\overset{−} {+}\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} ,\overset{−} {+}\:\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$ $$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$ $${Res}\left(\varphi,{i}\right)=\:\:\frac{−\mathrm{1}}{\mathrm{2}{ix}^{\mathrm{2}} \left(\mathrm{1}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:=\:\frac{−\mathrm{1}}{\mathrm{2}{i}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$ $$ \\ $$

Commented byprof Abdo imad last updated on 23/Jun/18

Res(ϕ,(1/(√x))e^(i(π/4)) )= (i/(x^3 ((i/x)+1)((2/(√x))e^((iπ)/4) )(((2i)/x)))) = ((√x)/(x^2 ((i/x)+1)4 e^((iπ)/4) )) = (((√x) e^(−((iπ)/4)) )/(4(xi +x^2 ))) Res(ϕ,−(1/(√x))e^(−((iπ)/4)) ) = ((−i)/(x^3 (−(i/x)+1)(((−2i)/x))(((−2)/(√x))e^(−((iπ)/4)) ))) = ((−(√x))/(x^2 (−(i/x)+1)4 e^(−((iπ)/4)) )) = ((−(√x)e^((iπ)/4) )/(4(x^2 −ix))) ∫_(−∞) ^(+∞) f(z)dz=2iπ{ ((−1)/(2i(x^2 +1))) + (((√x)e^(−((iπ)/4)) )/(4x(x+i))) −(((√x)e^((iπ)/4) )/(4x(x−i)))} =((−π)/(x^2 +1)) +((iπ(√x))/(2x)){ (e^(−((iπ)/4)) /(x+i)) −(e^((iπ)/4) /(x−i))} =((−π)/(x^2 +1)) +((iπ(√x))/(2x)){ (((x−i)e^(−((iπ)/4)) −(x+i)e^((iπ)/4) )/(x^2 +1))} =−(π/(x^2 +1)) −((iπ(√x))/(2x(x^2 +1))) 2iIm((x+i)e^((iπ)/4) ) =−(π/(x^2 +1)) +((π(√x))/(x(x^2 +1)))Im{(x+i)e^((iπ)/4) } but (x+i)e^((iπ)/4) =(x+i)((1/(√2)) +(1/(√2))i) =(1/(√2))(x+i)(1+i)=(1/(√2))( x +xi +i −1) =(1/(√2)){ x−1 +i(x+1)} ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =−(π/(x^2 +1)) +((π(√x))/((√2)x(x^2 +1)))(x+1)=f^′ (x) ⇒ f(x)=−π arctan(x) +(π/(√2))∫_. ^x (((t+1)(√t))/(t(t^2 +1)))dt +c

$${Res}\left(\varphi,\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)=\:\:\frac{{i}}{{x}^{\mathrm{3}} \left(\frac{{i}}{{x}}+\mathrm{1}\right)\left(\frac{\mathrm{2}}{\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\frac{\mathrm{2}{i}}{{x}}\right)} \\ $$ $$=\:\frac{\sqrt{{x}}}{{x}^{\mathrm{2}} \left(\frac{{i}}{{x}}+\mathrm{1}\right)\mathrm{4}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:=\:\frac{\sqrt{{x}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\left({xi}\:+{x}^{\mathrm{2}} \right)} \\ $$ $${Res}\left(\varphi,−\frac{\mathrm{1}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\:\:\:\frac{−{i}}{{x}^{\mathrm{3}} \left(−\frac{{i}}{{x}}+\mathrm{1}\right)\left(\frac{−\mathrm{2}{i}}{{x}}\right)\left(\frac{−\mathrm{2}}{\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$ $$=\:\:\frac{−\sqrt{{x}}}{{x}^{\mathrm{2}} \left(−\frac{{i}}{{x}}+\mathrm{1}\right)\mathrm{4}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:=\:\frac{−\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}\left({x}^{\mathrm{2}} −{ix}\right)} \\ $$ $$\int_{−\infty} ^{+\infty} \:{f}\left({z}\right){dz}=\mathrm{2}{i}\pi\left\{\:\:\frac{−\mathrm{1}}{\mathrm{2}{i}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\:\frac{\sqrt{{x}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{x}\left({x}+{i}\right)}\:−\frac{\sqrt{{x}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{x}\left({x}−{i}\right)}\right\} \\ $$ $$=\frac{−\pi}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\frac{{i}\pi\sqrt{{x}}}{\mathrm{2}{x}}\left\{\:\:\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{{x}+{i}}\:−\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{{x}−{i}}\right\} \\ $$ $$=\frac{−\pi}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\frac{{i}\pi\sqrt{{x}}}{\mathrm{2}{x}}\left\{\:\frac{\left({x}−{i}\right){e}^{−\frac{{i}\pi}{\mathrm{4}}} \:−\left({x}+{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} }{{x}^{\mathrm{2}} \:+\mathrm{1}}\right\} \\ $$ $$=−\frac{\pi}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{i}\pi\sqrt{{x}}}{\mathrm{2}{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\mathrm{2}{iIm}\left(\left({x}+{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} \right) \\ $$ $$=−\frac{\pi}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\:+\frac{\pi\sqrt{{x}}}{{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{Im}\left\{\left({x}+{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} \right\}\:{but} \\ $$ $$\left({x}+{i}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} =\left({x}+{i}\right)\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{i}\right) \\ $$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left({x}+{i}\right)\left(\mathrm{1}+{i}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(\:{x}\:+{xi}\:+{i}\:−\mathrm{1}\right) \\ $$ $$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left\{\:{x}−\mathrm{1}\:+{i}\left({x}+\mathrm{1}\right)\right\}\:\Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=−\frac{\pi}{{x}^{\mathrm{2}} +\mathrm{1}}\:\:+\frac{\pi\sqrt{{x}}}{\sqrt{\mathrm{2}}{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\left({x}+\mathrm{1}\right)={f}^{'} \left({x}\right) \\ $$ $$\Rightarrow\:{f}\left({x}\right)=−\pi\:{arctan}\left({x}\right)\:+\frac{\pi}{\sqrt{\mathrm{2}}}\int_{.} ^{{x}} \:\frac{\left({t}+\mathrm{1}\right)\sqrt{{t}}}{{t}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt}\:\:+{c} \\ $$ $$ \\ $$

Commented byprof Abdo imad last updated on 23/Jun/18

changement (√t)=u give ∫_. ^x (((t+1)(√t))/(t(t^2 +1)))dt = ∫_. ^(√x) (((u^2 +1)u)/(u^2 (u^4 +1))) 2u du =∫_. ^(√x) ((2(1+u^2 ))/(1+u^4 )) du...be cpntinued...

$${changement}\:\sqrt{{t}}={u}\:{give} \\ $$ $$\int_{.} ^{{x}} \:\:\:\frac{\left({t}+\mathrm{1}\right)\sqrt{{t}}}{{t}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:=\:\int_{.} ^{\sqrt{{x}}} \:\frac{\left({u}^{\mathrm{2}} +\mathrm{1}\right){u}}{{u}^{\mathrm{2}} \left({u}^{\mathrm{4}} +\mathrm{1}\right)}\:\mathrm{2}{u}\:{du} \\ $$ $$=\int_{.} ^{\sqrt{{x}}} \:\:\:\:\frac{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{\mathrm{1}+{u}^{\mathrm{4}} }\:{du}...{be}\:{cpntinued}... \\ $$

Commented bymath khazana by abdo last updated on 26/Jun/18

let drcompose F(u)= ((2u^2 +2)/(u^4 +1)) F(u)= ((2u^2 +2)/((u^2 +1)^2 −2u^2 )) =((2u^2 +2)/((u^2 +(√2) u+1)(u^2 −(√2) u+1))) =((au +b)/(u^2 +(√2)u +1)) +((cu +d)/(u^2 −(√2)u +1)) F(−u)=F(u) ⇒((−au +b)/(u^2 −(√2)u +1)) +((−cu +d)/(u^2 +(√2)u+1))=F(u) ⇒c=−a and b=d ⇒ F(u)= ((au+b)/(u^2 +(√2)u +1)) +((−au +b)/(u^2 −(√2)u +1)) F(0)=2= 2b ⇒b=1 ⇒ F(u)= ((au +1)/(u^2 +(√2)u+1)) +((−au +1)/(u^2 −(√2)u +1)) F(1)=2 =((a+1)/(2+(√2))) +((−a+1)/(2−(√2))) =(((2−(√2))a +2−(√2) −(2+(√2))a +2+(√2))/2) =((−2(√2)a +4)/2) =−(√2)a +2=2 ⇒a=0 ⇒ F(u)= (1/(u^2 +(√2)u +1)) +(1/(u^2 −(√2) u +1)) ∫ F(u) du = ∫ (du/(u^2 +2((√2)/2)u +(1/2) +(1/2))) + ∫ (du/(u^2 −2((√2)/2)u +(1/2) +(1/2))) =∫ (du/((u +(1/(√2)))^2 +(1/2))) +∫ (du/((u−(1/(√2)))^2 +(1/2))) =I +J I =_(u+(1/(√2))=(1/(√2))t) ∫ (1/((1/2)(t^2 +1))) (dt/(√2)) =(√2) arctan(u(√2) +1) J=_(u−(1/(√2))= (1/(√2))t) (√2) arctan(u(√2)−1) ⇒ ∫_. ^(√x) F(u)du = (√2)arctan((√(2x)) +1)+(√2)arctan((√(2x))−1)

$${let}\:{drcompose}\:{F}\left({u}\right)=\:\frac{\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{2}}{{u}^{\mathrm{4}} \:+\mathrm{1}} \\ $$ $${F}\left({u}\right)=\:\frac{\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{2}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}{u}^{\mathrm{2}} }\:=\frac{\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{2}}{\left({u}^{\mathrm{2}} +\sqrt{\mathrm{2}}\:{u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{u}+\mathrm{1}\right)} \\ $$ $$\:=\frac{{au}\:+{b}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{u}\:+\mathrm{1}} \\ $$ $${F}\left(−{u}\right)={F}\left({u}\right)\:\Rightarrow\frac{−{au}\:+{b}}{{u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:+\frac{−{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}+\mathrm{1}}={F}\left({u}\right) \\ $$ $$\Rightarrow{c}=−{a}\:{and}\:{b}={d}\:\Rightarrow \\ $$ $${F}\left({u}\right)=\:\frac{{au}+{b}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:+\frac{−{au}\:+{b}}{{u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}{u}\:+\mathrm{1}} \\ $$ $${F}\left(\mathrm{0}\right)=\mathrm{2}=\:\mathrm{2}{b}\:\Rightarrow{b}=\mathrm{1}\:\Rightarrow \\ $$ $${F}\left({u}\right)=\:\frac{{au}\:+\mathrm{1}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}+\mathrm{1}}\:+\frac{−{au}\:+\mathrm{1}}{{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}{u}\:+\mathrm{1}} \\ $$ $${F}\left(\mathrm{1}\right)=\mathrm{2}\:=\frac{{a}+\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:+\frac{−{a}+\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{2}}} \\ $$ $$=\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){a}\:+\mathrm{2}−\sqrt{\mathrm{2}}\:−\left(\mathrm{2}+\sqrt{\mathrm{2}}\right){a}\:+\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$ $$=\frac{−\mathrm{2}\sqrt{\mathrm{2}}{a}\:\:+\mathrm{4}}{\mathrm{2}}\:=−\sqrt{\mathrm{2}}{a}\:+\mathrm{2}=\mathrm{2}\:\Rightarrow{a}=\mathrm{0}\:\Rightarrow \\ $$ $${F}\left({u}\right)=\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}\:+\mathrm{1}}\:+\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:−\sqrt{\mathrm{2}}\:{u}\:+\mathrm{1}} \\ $$ $$\int\:{F}\left({u}\right)\:{du}\:=\:\int\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\:\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ $$+\:\int\:\:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:−\mathrm{2}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{u}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ $$=\int\:\:\:\:\frac{{du}}{\left({u}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:+\int\:\:\:\:\:\frac{{du}}{\left({u}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$ $$={I}\:+{J} \\ $$ $${I}\:=_{{u}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{t}} \:\:\:\:\:\int\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\frac{{dt}}{\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}}\:{arctan}\left({u}\sqrt{\mathrm{2}}\:+\mathrm{1}\right) \\ $$ $${J}=_{{u}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{t}} \:\:\:\sqrt{\mathrm{2}}\:{arctan}\left({u}\sqrt{\mathrm{2}}−\mathrm{1}\right)\:\Rightarrow \\ $$ $$\int_{.} ^{\sqrt{{x}}} \:{F}\left({u}\right){du}\:=\:\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}{x}}\:+\mathrm{1}\right)+\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}{x}}−\mathrm{1}\right) \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$

Commented bymath khazana by abdo last updated on 26/Jun/18

⇒F(x)=−π arctan(x) +π arctan((√(2x)) +1) +π arctan((√(2x)) −1) +λ λ= lim_(x→0) F(x)=0 ⇒ F(x)=π{ arctan((√(2x))+1)+arctan((√(2x))−1) −arctanx}

$$\Rightarrow{F}\left({x}\right)=−\pi\:{arctan}\left({x}\right)\:+\pi\:{arctan}\left(\sqrt{\mathrm{2}{x}}\:+\mathrm{1}\right) \\ $$ $$+\pi\:{arctan}\left(\sqrt{\mathrm{2}{x}}\:−\mathrm{1}\right)\:+\lambda \\ $$ $$\lambda=\:{lim}_{{x}\rightarrow\mathrm{0}} {F}\left({x}\right)=\mathrm{0}\:\Rightarrow \\ $$ $${F}\left({x}\right)=\pi\left\{\:{arctan}\left(\sqrt{\mathrm{2}{x}}+\mathrm{1}\right)+{arctan}\left(\sqrt{\mathrm{2}{x}}−\mathrm{1}\right)\:−{arctanx}\right\} \\ $$

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