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Question Number 39389 by maxmathsup by imad last updated on 05/Jul/18
calculate F(x) = ∫_0 ^∞ (dt/(1+(1+x(1+t^2 ))^2 ))
$${calculate}\:{F}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dt}}{\mathrm{1}+\left(\mathrm{1}+{x}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right)^{\mathrm{2}} } \\ $$
Commented by prof Abdo imad last updated on 24/Jul/18
changement t=tanθ give F(x) = ∫_0 ^(π/2) (((1+tan^2 θ))/(1+{1+x(1+tan^2 θ)}^2 ))dθ = ∫_0 ^(π/2) (1/(cos^2 θ{1+(x/(cos^2 θ))}^2 ))dθ = ∫_0 ^(π/2) ((cos^2 θ)/({x +cos^2 θ}^2 ))dθ =∫_0 ^(π/2) ((x+cos^2 θ−x)/((x+cos^2 θ)^2 ))dθ = ∫_0 ^(π/2) (dθ/(x +cos^2 θ)) −x ∫_0 ^(π/2) (dθ/((x+cos^2 θ)^2 )) but ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) = ∫_0 ^(π/2) (dθ/(x+((1+cos(2θ))/2))) = ∫_0 ^(π/2) ((2dθ)/(2x+1+cos(2θ))) =_(2θ=t) ∫_0 ^π (dt/(2x+1+cos(t))) =_(u=tan((t/2))) ∫_0 ^∞ (1/(2x+1 +((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫_0 ^∞ ((2du)/((2x+1)u^2 +1−u^2 )) = ∫_0 ^∞ ((2du)/(2xu^2 +1)) if x>0 ∫_0 ^∞ ((2du)/(1+2xu^2 )) =_((√(2x))u=α) ∫_0 ^∞ (2/(1+α^2 )) (dα/( (√(2x)))) =((√2)/( (√x))) .(π/2) if x<0 ∫_0 ^∞ ((2du)/(1+2xu^2 )) = ∫_0 ^∞ ((2du)/(1−((√(−2x))u)^2 )) =∫_0 ^∞ ((2du)/((1−(√(2x))u)(1+(√(2x))u))) = ∫_0 ^∞ { (1/(1−(√(2x))u)) +(1/(1+(√(2x))u))}du =[ln∣((1+(√(2x))u)/(1−(√(2x))u))∣]_0 ^(+∞) =0 so ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) =((π(√2))/( (√(2x)))) if x>0 ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) =0 if x<0
$${changement}\:{t}={tan}\theta\:{give} \\ $$$${F}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{\mathrm{1}+\left\{\mathrm{1}+{x}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)\right\}^{\mathrm{2}} }{d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta\left\{\mathrm{1}+\frac{{x}}{{cos}^{\mathrm{2}} \theta}\right\}^{\mathrm{2}} }{d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cos}^{\mathrm{2}} \theta}{\left\{{x}\:+{cos}^{\mathrm{2}} \theta\right\}^{\mathrm{2}} }{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{x}+{cos}^{\mathrm{2}} \theta−{x}}{\left({x}+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }{d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{{x}\:+{cos}^{\mathrm{2}} \theta}\:−{x}\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{\left({x}+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:{but} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{{x}+{cos}^{\mathrm{2}} \theta}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{{x}+\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{2}{d}\theta}{\mathrm{2}{x}+\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}\:=_{\mathrm{2}\theta={t}} \:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}{x}+\mathrm{1}+{cos}\left({t}\right)} \\ $$$$=_{{u}={tan}\left(\frac{{t}}{\mathrm{2}}\right)} \:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}\:+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{2}{x}+\mathrm{1}\right){u}^{\mathrm{2}} \:+\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}{xu}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${if}\:{x}>\mathrm{0}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{du}}{\mathrm{1}+\mathrm{2}{xu}^{\mathrm{2}} }\:=_{\sqrt{\mathrm{2}{x}}{u}=\alpha} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\frac{{d}\alpha}{\:\sqrt{\mathrm{2}{x}}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{{x}}}\:.\frac{\pi}{\mathrm{2}} \\ $$$${if}\:{x}<\mathrm{0}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+\mathrm{2}{xu}^{\mathrm{2}} }\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{du}}{\mathrm{1}−\left(\sqrt{−\mathrm{2}{x}}{u}\right)^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}−\sqrt{\mathrm{2}{x}}{u}\right)\left(\mathrm{1}+\sqrt{\mathrm{2}{x}}{u}\right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\left\{\:\frac{\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{2}{x}}{u}}\:+\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}{x}}{u}}\right\}{du} \\ $$$$=\left[{ln}\mid\frac{\mathrm{1}+\sqrt{\mathrm{2}{x}}{u}}{\mathrm{1}−\sqrt{\mathrm{2}{x}}{u}}\mid\right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{0}\:\:{so} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{{x}+{cos}^{\mathrm{2}} \theta}\:=\frac{\pi\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}{x}}}\:{if}\:{x}>\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{{x}+{cos}^{\mathrm{2}} \theta}\:=\mathrm{0}\:{if}\:{x}<\mathrm{0} \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 24/Jul/18
let w(x)= ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) ⇒ w^′ (x) =−∫_0 ^(π/2) (dθ/((x +cos^2 θ)^2 )) ⇒ ∫_0 ^(π/2) (dθ/((x+cos^2 θ)^2 )) =−w^′ (x) so if x>0 w(x)=((π(√2))/(2(√x))) ⇒w^′ (x)=(π/( (√2))) (−(1/(2x(√x)))) =−(π/(2(√2)x(√x))) ⇒ ∫_0 ^(π/2) (dθ/((x+cos^2 θ)^2 )) = (π/(2(√2)x(√x))) and F(x)=((π(√2))/(2(√x))) −x (π/(2(√2)x(√x))) =((π(√2))/(2(√x))) −(π/(2(√2)(√x))) =(π/( (√x))){((√2)/2) −(1/(2(√2)))}=(π/( (√x))){(2/(4(√2)))} = (π/(2(√(2x)))) if x<0 w(x)=0 ⇒ ∫_0 ^(π/2) (dθ/((x+cos^2 θ)^2 )) =0 ⇒ F(x) = 0 because ∫_0 ^(π/2) (dθ/(x+cos^2 θ)) =0
$${let}\:{w}\left({x}\right)=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{{x}+{cos}^{\mathrm{2}} \theta}\:\Rightarrow \\ $$$${w}^{'} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{d}\theta}{\left({x}\:+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{\left({x}+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:=−{w}^{'} \left({x}\right)\:{so} \\ $$$${if}\:{x}>\mathrm{0}\:{w}\left({x}\right)=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{{x}}}\:\Rightarrow{w}^{'} \left({x}\right)=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{x}}}\right) \\ $$$$=−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}{x}\sqrt{{x}}}\:\Rightarrow\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{d}\theta}{\left({x}+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}{x}\sqrt{{x}}} \\ $$$${and}\:{F}\left({x}\right)=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{{x}}}\:−{x}\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}{x}\sqrt{{x}}} \\ $$$$=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{{x}}}\:−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}\sqrt{{x}}}\:=\frac{\pi}{\:\sqrt{{x}}}\left\{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right\}=\frac{\pi}{\:\sqrt{{x}}}\left\{\frac{\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}\right\} \\ $$$$=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}{x}}} \\ $$$${if}\:{x}<\mathrm{0}\:\:{w}\left({x}\right)=\mathrm{0}\:\Rightarrow\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{\left({x}+{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\mathrm{0}\:\:{because}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{{x}+{cos}^{\mathrm{2}} \theta}\:=\mathrm{0} \\ $$

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