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Question Number 128225 by mnjuly1970 last updated on 05/Jan/21
...nice calculus... prove that : φ = ∫_0 ^( π) ((tan^(−1) (tan^2 (x)))/(tan^2 (x)) dx=^(???) π((√2) −(π/4))
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\:\phi\:=\:\int_{\mathrm{0}} ^{\:\pi} \frac{{tan}^{−\mathrm{1}} \left({tan}^{\mathrm{2}} \left({x}\right)\right)}{{tan}^{\mathrm{2}} \left({x}\right.}\:{dx}\overset{???} {=}\pi\left(\sqrt{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 05/Jan/21
Φ=∫_0 ^π ((arctan(tan^2 x))/(tan^2 x))dx let f(a) =∫_0 ^π ((arctan(atan^2 x))/(tan^2 x))dx (a>0) ⇒ f^′ (a)=∫_0 ^π (1/((1+a^2 tan^4 x)))dx =∫_0 ^(π/2) (...)dx +∫_(π/2) ^π (...)dx ∫_0 ^(π/2) (dx/(1+a^2 tan^4 x)) =_(tanx=t) ∫_0 ^∞ (dt/((1+t^2 )(1+a^2 t^4 ))) =(1/2)∫_(−∞) ^(+∞) (dt/((t^2 +1)(a^2 t^4 +1))) let ϕ(z)=(1/((z^2 +1)(a^2 z^4 +1))) ⇒ϕ(z)=(1/(a^2 (z^2 +1)(z^4 +(1/a^2 )))) =(1/(a^2 (z−i)(z+i)(z^2 −(i/a))(z^2 +(i/a)))) =(1/(a^2 (z−i)(z+i)(z−(1/( (√a)))e^((iπ)/4) )(z+(1/( (√a)))e^((iπ)/4) )(z−(1/( (√a)))e^(−((iπ)/4)) )(z+(1/( (√a)))e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,i) +Res(ϕ,(1/( (√a)))e^((iπ)/4) ) +Res(ϕ,−(1/( (√a)))e^(−((iπ)/4)) )} ...be continued...
$$\Phi=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{arctan}\left(\mathrm{tan}^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{tan}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{arctan}\left(\mathrm{atan}^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{tan}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:\:\:\left(\mathrm{a}>\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \:\mathrm{tan}^{\mathrm{4}} \mathrm{x}\right)}\mathrm{dx}\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(…\right)\mathrm{dx}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:\:\left(…\right)\mathrm{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{tan}^{\mathrm{4}} \mathrm{x}}\:=_{\mathrm{tanx}=\mathrm{t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \:\mathrm{t}^{\mathrm{4}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{a}^{\mathrm{2}} \:\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}\right)}\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{a}^{\mathrm{2}} \mathrm{z}^{\mathrm{4}} \:+\mathrm{1}\right)} \\ $$$$\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{4}} \:+\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} −\frac{\mathrm{i}}{\mathrm{a}}\right)\left(\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{i}}{\mathrm{a}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)\left(\mathrm{z}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\:\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:+\mathrm{Res}\left(\varphi,\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:+\mathrm{Res}\left(\varphi,−\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$…\mathrm{be}\:\mathrm{continued}… \\ $$
Answered by mindispower last updated on 06/Jan/21
tan(x)=t =∫_(−∞) ^∞ ((tan^− (t^2 ))/(t^2 (1+t^2 )))dt ∫_(−∞) ^∞ ((tan^− (t^2 ))/t^2 )−∫_(−∞) ^∞ ((tan^− (t^2 ))/(1+t^2 ))dt =2(∫_0 ^∞ ((tan^− (t^2 ))/t^2 )−∫_0 ^∞ ((tan^− (t^2 ))/(1+t^2 )))dt =2A−2B A=[−((tan^− (t^2 ))/t)]_0 ^∞ +2∫_0 ^∞ (dt/(1+t^4 )) ∣t^4 =u =(1/2)∫_0 ^∞ (u^(−(3/4)) /(1+u))=(1/2)β((1/4),(3/4))=(π/(2sin((π/4))))=(π/( (√2))) B =[tan^− (t)tan^− (t^2 )]−∫_0 ^∞ ((2ttan^− (t))/(1+t^4 )) =(π^2 /4)−∫_0 ^∞ t^4 (((2/t)((π/2)−tan^− (t)))/(1+t^4 )).(dt/t^2 ) _(=(π^2 /4)−C) =(π^2 /4)−∫_0 ^∞ ((tπ)/(1+t^4 ))+∫_0 ^∞ ((2ttan^− (t))/(1+t^4 ))dt ⇒2C=∫_0 ^∞ ((πt)/(1+t^4 ))=(π/2)∫_0 ^∞ .((d(t^2 ))/(1+(t^2 )^2 ))=(π/(2.)).tan^− (t^2 )]_0 ^∞ =(π^2 /4)⇒C=(π^2 /8) B=(π^2 /8) 2A−2B=π(√2)−(π^2 /4)=π((√2)−(π/4))=∫_0 ^(π/2) ((tan^− (tg^2 (t)))/(tg^2 (t)))dt
$${tan}\left({x}\right)={t} \\ $$$$=\int_{−\infty} ^{\infty} \frac{{tan}^{−} \left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\int_{−\infty} ^{\infty} \frac{{tan}^{−} \left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }−\int_{−\infty} ^{\infty} \frac{{tan}^{−} \left({t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{2}\left(\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt} \\ $$$$=\mathrm{2}{A}−\mathrm{2}{B} \\ $$$${A}=\left[−\frac{{tan}^{−} \left({t}^{\mathrm{2}} \right)}{{t}}\right]_{\mathrm{0}} ^{\infty} +\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:\:\mid{t}^{\mathrm{4}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} }{\mathrm{1}+{u}}=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}}\right)}=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$$${B}\:=\left[{tan}^{−} \left({t}\right){tan}^{−} \left({t}^{\mathrm{2}} \right)\right]−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{ttan}^{−} \left({t}\right)}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\int_{\mathrm{0}} ^{\infty} {t}^{\mathrm{4}} \frac{\frac{\mathrm{2}}{{t}}\left(\frac{\pi}{\mathrm{2}}−{tan}^{−} \left({t}\right)\right)}{\mathrm{1}+{t}^{\mathrm{4}} }.\frac{{dt}}{{t}^{\mathrm{2}} }\:\:_{=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−{C}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\int_{\mathrm{0}} ^{\infty} \frac{{t}\pi}{\mathrm{1}+{t}^{\mathrm{4}} }+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{ttan}^{−} \left({t}\right)}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\left.\Rightarrow\mathrm{2}{C}=\int_{\mathrm{0}} ^{\infty} \frac{\pi{t}}{\mathrm{1}+{t}^{\mathrm{4}} }=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} .\frac{{d}\left({t}^{\mathrm{2}} \right)}{\mathrm{1}+\left({t}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}.}.{tan}^{−} \left({t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\Rightarrow{C}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${B}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\mathrm{2}{A}−\mathrm{2}{B}=\pi\sqrt{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}=\pi\left(\sqrt{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tan}^{−} \left({tg}^{\mathrm{2}} \left({t}\right)\right)}{{tg}^{\mathrm{2}} \left({t}\right)}{dt} \\ $$$$ \\ $$$$ \\ $$

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