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Question Number 86167 by jagoll last updated on 27/Mar/20

∫_0 ^(π/2) ((arc tan ((√(tan x))))/(tan x)) dx

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{arc}\:\mathrm{tan}\:\left(\sqrt{\mathrm{tan}\:\mathrm{x}}\right)}{\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$

Commented by mathmax by abdo last updated on 27/Mar/20

let take a try I =∫_0 ^(π/2) ((arctan((√(tanx))))/(tanx))dx changement (√(tanx))=t give tanx=t^2 ⇒ x =arctan(t^2 ) ⇒ I =∫_0 ^(+∞) ((arctan(t))/t^2 )×((2t)/(1+t^4 ))dt =2 ∫_0 ^∞ ((arctant)/(t(1+t^4 )))dt the convergence of this integral is assured let f(a) =∫_0 ^∞ ((arctan(at))/(t(1+t^4 )))dt with a>0 f^′ (a) =∫_0 ^∞ (t/(t(1+a^2 t^2 )(1+t^4 )))dt =∫_0 ^∞ (dt/((t^4 +1)(1+a^2 t^2 ))) =_(at=u) ∫_0 ^∞ (du/(a((u^4 /a^4 ) +1)(1+u^2 ))) =∫_0 ^∞ ((a^3 du)/((u^2 +1)(u^4 +a^4 ))) ⇒ 2f^′ (a) =a^3 ∫_(−∞) ^(+∞) (du/((u^2 +1)(u^4 +a^4 ))) let ϕ(z) =(1/((z^2 +1)(z^4 +a^4 ))) ⇒ ϕ(z) =(1/((z−i)(z+i)(z^2 −ia^2 )(z^2 +ia^2 ))) =(1/((z−i)(z+i)(z−ae^((iπ)/4) )(z+ae^((iπ)/4) )(z−ae^(−((iπ)/4)) )(z+a e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,i) +Res(ϕ,ae^((iπ)/4) ) +Re(ϕ,−ae^(−((iπ)/4)) )} Res(ϕ,i) =(1/((2i)(1+a^4 ))) Res(ϕ,ae^((iπ)/4) ) =(1/((a^2 i+1)(2ae^((iπ)/4) )(2ia^2 ))) =(e^(−((iπ)/4)) /(4ia^3 (1+a^2 i))) Res(ϕ,−ae^(−((iπ)/4)) ) =(1/((−a^2 i+1)(−2ia^2 )(−2ae^(−((iπ)/4)) ))) =(e^((iπ)/4) /(4ia^3 (1−a^2 i))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(1/((2i)(1+a^4 ))) +(e^(−((iπ)/4)) /(4ia^3 (1+a^2 i))) +(e^((iπ)/4) /(4ia^3 (1−a^2 i)))} =(π/(1+a^4 )) +(1/(2a^3 ))( (e^(−((iπ)/4)) /(1+a^2 i)) +(e^((iπ)/4) /(1−a^2 i))) =(π/(1+a^4 )) +(1/(2a^3 ))(2Re((e^((iπ)/4) /(1−a^2 i)))) =(π/(1+a^4 )) +(1/a^3 )Re((e^((iπ)/4) /(1−a^2 i))) (e^((iπ)/4) /(1−a^2 i)) =(((1+a^2 i) e^((iπ)/4) )/(1+a^4 )) =(((1+a^2 i)((1/(√2))+(i/(√2))))/(1+a^4 )) =(1/((√2)(1+a^4 )))(1+a^2 i)(1+i) =(1/((√2)(1+a^4 )))(1+i+a^2 i−a^2 ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =(π/(1+a^4 )) +((1−a^2 )/(a^3 (√2)(1+a^4 ))) ⇒ 2f^′ (a) =((πa^3 )/(1+a^4 )) +((1−a^2 )/((√2)(1+a^4 ))) ⇒f^′ (a) =(π/2)×(a^3 /(1+a^4 )) +(1/(2(√2)))×((1−a^2 )/(1+a^4 )) ⇒ f(a) =(π/2)∫_0 ^a (x^3 /(1+x^4 ))dx +(1/(2(√2))) ∫_0 ^a ((1−x^2 )/(1+x^4 ))dx +C C=f(0) =0 ⇒f(a) =(π/8)ln(1+a^4 )+(1/(2(√2)))∫_0 ^a ((1−x^2 )/(1+x^4 ))dx I =2f(1) ....be continued....

$${let}\:{take}\:{a}\:{try} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{arctan}\left(\sqrt{{tanx}}\right)}{{tanx}}{dx}\:{changement}\:\sqrt{{tanx}}={t}\:{give}\:{tanx}={t}^{\mathrm{2}} \Rightarrow \\ $$$${x}\:={arctan}\left({t}^{\mathrm{2}} \right)\:\Rightarrow\:{I}\:=\int_{\mathrm{0}} ^{+\infty} \:\frac{{arctan}\left({t}\right)}{{t}^{\mathrm{2}} }×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctant}}{{t}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{dt}\:\:{the}\:{convergence}\:{of}\:{this}\:{integral}\:{is}\:{assured} \\ $$$${let}\:\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({at}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{dt}\:\:{with}\:{a}>\mathrm{0} \\ $$$${f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}}{{t}\left(\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{4}} \:+\mathrm{1}\right)\left(\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} \right)} \\ $$$$=_{{at}={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{{a}\left(\frac{{u}^{\mathrm{4}} }{{a}^{\mathrm{4}} }\:+\mathrm{1}\right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{a}^{\mathrm{3}} \:{du}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{4}} \:+{a}^{\mathrm{4}} \right)}\:\Rightarrow \\ $$$$\mathrm{2}{f}^{'} \left({a}\right)\:={a}^{\mathrm{3}} \:\int_{−\infty} ^{+\infty} \:\frac{{du}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{4}} \:+{a}^{\mathrm{4}} \right)}\:{let}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{4}} \:+{a}^{\mathrm{4}} \right)}\:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}^{\mathrm{2}} −{ia}^{\mathrm{2}} \right)\left({z}^{\mathrm{2}} \:+{ia}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{ae}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{a}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Re}\left(\varphi,−{ae}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,{i}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)\left(\mathrm{1}+{a}^{\mathrm{4}} \right)} \\ $$$${Res}\left(\varphi,{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} {i}+\mathrm{1}\right)\left(\mathrm{2}{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{2}{ia}^{\mathrm{2}} \right)}\:=\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{ia}^{\mathrm{3}} \left(\mathrm{1}+{a}^{\mathrm{2}} {i}\right)} \\ $$$${Res}\left(\varphi,−{ae}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{1}}{\left(−{a}^{\mathrm{2}} {i}+\mathrm{1}\right)\left(−\mathrm{2}{ia}^{\mathrm{2}} \right)\left(−\mathrm{2}{ae}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:=\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{ia}^{\mathrm{3}} \left(\mathrm{1}−{a}^{\mathrm{2}} {i}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)\left(\mathrm{1}+{a}^{\mathrm{4}} \right)}\:+\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{ia}^{\mathrm{3}} \left(\mathrm{1}+{a}^{\mathrm{2}} {i}\right)}\:+\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{4}{ia}^{\mathrm{3}} \left(\mathrm{1}−{a}^{\mathrm{2}} {i}\right)}\right\} \\ $$$$=\frac{\pi}{\mathrm{1}+{a}^{\mathrm{4}} }\:+\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{3}} }\left(\:\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{1}+{a}^{\mathrm{2}} {i}}\:+\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{1}−{a}^{\mathrm{2}} {i}}\right) \\ $$$$=\frac{\pi}{\mathrm{1}+{a}^{\mathrm{4}} }\:+\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{3}} }\left(\mathrm{2}{Re}\left(\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{1}−{a}^{\mathrm{2}} {i}}\right)\right)\:=\frac{\pi}{\mathrm{1}+{a}^{\mathrm{4}} }\:+\frac{\mathrm{1}}{{a}^{\mathrm{3}} }{Re}\left(\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{1}−{a}^{\mathrm{2}} {i}}\right) \\ $$$$\frac{{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{1}−{a}^{\mathrm{2}} {i}}\:=\frac{\left(\mathrm{1}+{a}^{\mathrm{2}} {i}\right)\:{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\mathrm{1}+{a}^{\mathrm{4}} }\:=\frac{\left(\mathrm{1}+{a}^{\mathrm{2}} {i}\right)\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{{i}}{\sqrt{\mathrm{2}}}\right)}{\mathrm{1}+{a}^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\left(\mathrm{1}+{a}^{\mathrm{4}} \right)}\left(\mathrm{1}+{a}^{\mathrm{2}} {i}\right)\left(\mathrm{1}+{i}\right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\left(\mathrm{1}+{a}^{\mathrm{4}} \right)}\left(\mathrm{1}+{i}+{a}^{\mathrm{2}} {i}−{a}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\pi}{\mathrm{1}+{a}^{\mathrm{4}} }\:+\frac{\mathrm{1}−{a}^{\mathrm{2}} }{{a}^{\mathrm{3}} \sqrt{\mathrm{2}}\left(\mathrm{1}+{a}^{\mathrm{4}} \right)}\:\Rightarrow \\ $$$$\mathrm{2}{f}^{'} \left({a}\right)\:=\frac{\pi{a}^{\mathrm{3}} }{\mathrm{1}+{a}^{\mathrm{4}} }\:+\frac{\mathrm{1}−{a}^{\mathrm{2}} }{\sqrt{\mathrm{2}}\left(\mathrm{1}+{a}^{\mathrm{4}} \right)}\:\Rightarrow{f}^{'} \left({a}\right)\:=\frac{\pi}{\mathrm{2}}×\frac{{a}^{\mathrm{3}} }{\mathrm{1}+{a}^{\mathrm{4}} }\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\mathrm{1}−{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{4}} }\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{{a}} \:\:\frac{{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{{a}} \:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:+{C} \\ $$$${C}={f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow{f}\left({a}\right)\:=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{1}+{a}^{\mathrm{4}} \right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{{a}} \:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\: \\ $$$${I}\:=\mathrm{2}{f}\left(\mathrm{1}\right)\:....{be}\:{continued}.... \\ $$

Commented by mathmax by abdo last updated on 27/Mar/20

let try anther way we have I =2∫_0 ^∞ ((arctant)/(t(1+t^4 )))dt ⇒I =∫_(−∞) ^(+∞) ((arctant)/(t(t^4 +1)))dt let ϕ(z) =((arctanz)/(z(z^4 +1))) ⇒ ϕ(z) =((arctanz)/(z(z^2 −i)(z^2 +i))) =((arctan(z))/(z(z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,0) +Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,0)=0 Res(ϕ,e^((iπ)/4) ) =((arctan(e^((iπ)/4) ))/(e^((iπ)/4) (2e^((iπ)/4) )(2i))) =((arctan(e^((iπ)/4) ))/(4i (i))) =−(1/4) arctan(e^((iπ)/4) ) Res(ϕ,−e^(−((iπ)/4)) ) =((−arctan(e^(−((iπ)/4)) ))/((−e^(−((iπ)/4)) )(−2e^(−((iπ)/4)) )(−2i))) =((arctan(e^(−((iπ)/4)) ))/(4i(−i))) =(1/4) arctan(e^(−((iπ)/4)) ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{−(1/4)arctan(e^((iπ)/4) )+(1/4) arctan(e^(−((iπ)/4)) )} =−((iπ)/2){ arctan(e^((iπ)/4) )−arctan(e^(−((iπ)/4)) )} we know arctan(z) =(1/(2i))ln(((1+iz)/(1−iz))) and arctan(z^− )=(1/(2i))ln(((1+iz^− )/(1−iz^− ))) ⇒arctan(z)−arctan(z^− ) =(1/(2i))ln(((1+iz)/(1−iz))×((1−iz^− )/(1+iz^− ))) =(1/(2i))ln(((1−iz^− +iz +zz^− )/(1+iz^− −iz +zz^− ))) =(1/(2i))ln(((1+i(2iIm(z)+∣z∣^2 )/(1−i(2iIm(z))+∣z∣^2 ))) =(1/(2i))ln(((1−2Im(z)+∣z∣^2 )/(1+2Im(z)+∣z∣^2 ))) ⇒ arctan(e^((iπ)/4) )−arctan(e^(−((iπ)/4)) ) =(1/(2i))ln(((1−2×((√2)/2)+1)/(1+2×((√2)/2)+1))) =(1/(2i))ln(((2−(√2))/(2+(√2)))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =−((iπ)/2)×(1/(2i))ln(((2−(√2))/(2+(√2)))) =−(π/4)ln(((2−(√2))/(2+(√2)))) =(π/4)ln(((2+(√2))/(2−(√2)))) =I

$${let}\:{try}\:{anther}\:{way}\:{we}\:{have}\:\:{I}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctant}}{{t}\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}{dt} \\ $$$$\Rightarrow{I}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{arctant}}{{t}\left({t}^{\mathrm{4}} \:+\mathrm{1}\right)}{dt}\:\:{let}\:\varphi\left({z}\right)\:=\frac{{arctanz}}{{z}\left({z}^{\mathrm{4}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\frac{{arctanz}}{{z}\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} \:+{i}\right)}\:=\frac{{arctan}\left({z}\right)}{{z}\left({z}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\mathrm{0}\right)\:+{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,\mathrm{0}\right)=\mathrm{0} \\ $$$${Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{{arctan}\left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}{{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{2}{i}\right)}\:=\frac{{arctan}\left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}{\mathrm{4}{i}\:\left({i}\right)}\:=−\frac{\mathrm{1}}{\mathrm{4}}\:{arctan}\left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right) \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\frac{−{arctan}\left({e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}{\left(−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left(−\mathrm{2}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left(−\mathrm{2}{i}\right)}\:=\frac{{arctan}\left({e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}{\mathrm{4}{i}\left(−{i}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:{arctan}\left({e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{−\frac{\mathrm{1}}{\mathrm{4}}{arctan}\left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right)+\frac{\mathrm{1}}{\mathrm{4}}\:{arctan}\left({e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$=−\frac{{i}\pi}{\mathrm{2}}\left\{\:{arctan}\left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right)−{arctan}\left({e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${we}\:{know}\:\:{arctan}\left({z}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{iz}}{\mathrm{1}−{iz}}\right)\:{and}\:{arctan}\left(\overset{−} {{z}}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{i}\overset{−} {{z}}}{\mathrm{1}−{i}\overset{−} {{z}}}\right) \\ $$$$\Rightarrow{arctan}\left({z}\right)−{arctan}\left(\overset{−} {{z}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{iz}}{\mathrm{1}−{iz}}×\frac{\mathrm{1}−{i}\overset{−} {{z}}}{\mathrm{1}+{i}\overset{−} {{z}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}−{i}\overset{−} {{z}}+{iz}\:+{z}\overset{−} {{z}}}{\mathrm{1}+{i}\overset{−} {{z}}−{iz}\:+{z}\overset{−} {{z}}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{i}\left(\mathrm{2}{iIm}\left({z}\right)+\mid{z}\mid^{\mathrm{2}} \right.}{\mathrm{1}−{i}\left(\mathrm{2}{iIm}\left({z}\right)\right)+\mid{z}\mid^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}−\mathrm{2}{Im}\left({z}\right)+\mid{z}\mid^{\mathrm{2}} }{\mathrm{1}+\mathrm{2}{Im}\left({z}\right)+\mid{z}\mid^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${arctan}\left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right)−{arctan}\left({e}^{−\frac{{i}\pi}{\mathrm{4}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}−\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{1}}{\mathrm{1}+\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\mathrm{1}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=−\frac{{i}\pi}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right)\:=−\frac{\pi}{\mathrm{4}}{ln}\left(\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\right) \\ $$$$=\frac{\pi}{\mathrm{4}}{ln}\left(\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\right)\:={I}\: \\ $$

Commented by M±th+et£s last updated on 28/Mar/20

∫_0 ^(π/2) ((tan^(−1) ((√(tan(x)))))/(tan(x))) dx y^2 =tan(x) tan^− (y^2 )=x dx=((2y)/(1+y^4 )) dy I=∫_0 ^∞ ((tan^(−1) (y))/y^2 ) . ((2y)/(1+y^4 )) dy=∫_0 ^∞ ((2 tan^(−1) (y))/(y(1+y^4 )))dy I(a)=∫_0 ^∞ ((2tan^(−1) (ay))/(y(1+y^4 )))dy I(a)=∫_0 ^∞ (((2y)/(1+a^2 y^2 ))/(y(1+y^4 )))dy=∫_0 ^∞ (2/((1+a^2 y^2 )(1+y^4 )))dy =(2/(a^4 +1))∫_0 ^∞ ((a^2 /(1+a^2 y^2 ))+((1−a^2 y^2 )/(1+y^4 )))dy =(2/(a^4 +1))[∫_0 ^∞ (a^2 /(1+a^2 y^2 ))dy + ∫_0 ^∞ (1/(1+y^4 ))dy −a^2 ∫_0 ^∞ (y^2 /(1+y^4 ))dy] (2/(a^4 +1))[((a^3 π)/2)+(π/(2(√2)))−((a^2 π)/(2(√2)))] ((2π)/(a^4 +1))[((a^4 −(√2) a^2 +(√2))/4)]=(π/2)[((2a^3 −(√2)a^2 +(√2))/((a^2 +1)^2 −2a^2 ))] =(π/2)[(((2a+(√2))(a^2 −(√2)a+1))/((a^2 +(√2)a+1)(a^2 −(√2)a+1)))] I(0)=∫_0 ^∞ ((2tan^(−1) (0))/(y(1+y^4 )))dy=0 ⇒⇒0=(π/2)ln(0+0+1)+c c=0 I(a)=(π/2)ln(a^2 +(√2) a+1) I(1)=(π/2)ln(2+(√2)) notice// ∫_0 ^∞ (dy/(1+y^4 ))=(π/(2(√2))) and ∫_0 ^∞ (y^2 /(1+y^4 ))dy=(π/(2(√2))) and ∫_0 ^∞ (dy/(1+a^2 y^2 ))=((aπ)/2)

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{tan}^{−\mathrm{1}} \left(\sqrt{{tan}\left({x}\right)}\right)}{{tan}\left({x}\right)}\:{dx} \\ $$$${y}^{\mathrm{2}} ={tan}\left({x}\right)\:\:\:{tan}^{−} \left({y}^{\mathrm{2}} \right)={x}\:\:{dx}=\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{4}} }\:{dy} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−\mathrm{1}} \left({y}\right)}{{y}^{\mathrm{2}} }\:.\:\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{4}} }\:{dy}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}\:{tan}^{−\mathrm{1}} \left({y}\right)}{{y}\left(\mathrm{1}+{y}^{\mathrm{4}} \right)}{dy} \\ $$$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{tan}^{−\mathrm{1}} \left({ay}\right)}{{y}\left(\mathrm{1}+{y}^{\mathrm{4}} \right)}{dy} \\ $$$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\frac{\mathrm{2}{y}}{\mathrm{1}+{a}^{\mathrm{2}} {y}^{\mathrm{2}} }}{{y}\left(\mathrm{1}+{y}^{\mathrm{4}} \right)}{dy}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}}{\left(\mathrm{1}+{a}^{\mathrm{2}} {y}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{4}} \right)}{dy} \\ $$$$=\frac{\mathrm{2}}{{a}^{\mathrm{4}} +\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left(\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} {y}^{\mathrm{2}} }+\frac{\mathrm{1}−{a}^{\mathrm{2}} {y}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{4}} }\right){dy} \\ $$$$=\frac{\mathrm{2}}{{a}^{\mathrm{4}} +\mathrm{1}}\left[\int_{\mathrm{0}} ^{\infty} \frac{{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} {y}^{\mathrm{2}} }{dy}\:+\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{4}} }{dy}\:−{a}^{\mathrm{2}} \int_{\mathrm{0}} ^{\infty} \frac{{y}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{4}} }{dy}\right] \\ $$$$\frac{\mathrm{2}}{{a}^{\mathrm{4}} +\mathrm{1}}\left[\frac{{a}^{\mathrm{3}} \pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{{a}^{\mathrm{2}} \pi}{\mathrm{2}\sqrt{\mathrm{2}}}\right] \\ $$$$\frac{\mathrm{2}\pi}{{a}^{\mathrm{4}} +\mathrm{1}}\left[\frac{{a}^{\mathrm{4}} −\sqrt{\mathrm{2}}\:{a}^{\mathrm{2}} +\sqrt{\mathrm{2}}}{\mathrm{4}}\right]=\frac{\pi}{\mathrm{2}}\left[\frac{\mathrm{2}{a}^{\mathrm{3}} −\sqrt{\mathrm{2}}{a}^{\mathrm{2}} +\sqrt{\mathrm{2}}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} }\right] \\ $$$$=\frac{\pi}{\mathrm{2}}\left[\frac{\left(\mathrm{2}{a}+\sqrt{\mathrm{2}}\right)\left({a}^{\mathrm{2}} −\sqrt{\mathrm{2}}{a}+\mathrm{1}\right)}{\left({a}^{\mathrm{2}} +\sqrt{\mathrm{2}}{a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} −\sqrt{\mathrm{2}}{a}+\mathrm{1}\right)}\right] \\ $$$${I}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{tan}^{−\mathrm{1}} \left(\mathrm{0}\right)}{{y}\left(\mathrm{1}+{y}^{\mathrm{4}} \right)}{dy}=\mathrm{0}\:\Rightarrow\Rightarrow\mathrm{0}=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{0}+\mathrm{0}+\mathrm{1}\right)+{c}\:\:\:{c}=\mathrm{0} \\ $$$${I}\left({a}\right)=\frac{\pi}{\mathrm{2}}{ln}\left({a}^{\mathrm{2}} +\sqrt{\mathrm{2}}\:{a}+\mathrm{1}\right) \\ $$$${I}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\:\: \\ $$$$ \\ $$$$ \\ $$$${notice}//\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{dy}}{\mathrm{1}+{y}^{\mathrm{4}} }=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \frac{{y}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{4}} }{dy}=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \frac{{dy}}{\mathrm{1}+{a}^{\mathrm{2}} {y}^{\mathrm{2}} }=\frac{{a}\pi}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by TANMAY PANACEA. last updated on 27/Mar/20

(√(tanx)) =tant tanx=tan^2 t sec^2 xdx=2tant.sec^2 tdt dx=((2tantsec^2 t)/(1+tan^4 t))dt ∫_0 ^(π/2) ((2tantsec^2 t)/(1+tan^4 t))×(t/(tan^2 t))dt (I/2)=∫_0 ^(π/2) t×(1/(cos^2 t×((sint)/(cost))))×((cos^4 t)/(sin^4 t+cos^4 t))dt (I/2)=∫_0 ^(π/2) ((π/2)−t)×(1/(sintcost))×((sin^4 t)/(cos^4 t+sin^4 t))dt using ∫_0 ^a f(x)dx=∫_0 ^a f(a−x)dx (I/2)+(I/2)=∫_0 ^(π/2) (π/2)×((sin^4 t)/(sintcost))×(1/(sin^4 t+cos^4 t)) ((2I)/π)=∫_0 ^(π/2) (1/(tant))×sec^2 t×(1/(1+tan^4 t))×dt a=tant ((2I)/π)=∫_0 ^∞ (da/(a(1+a^4 ))) =(1/2)∫_0 ^∞ ((d(a^2 ))/(a^2 (1+a^4 ))) replace a^2 by b ((4I)/π)=∫_0 ^∞ (db/(b(1+b^2 )))=(1/2)∫_0 ^∞ ((2bdb)/(b^2 (1+b^2 ))) ((8I)/π)=∫_0 ^∞ (dk/(k(1+k))) [k=b^2 ] =∫_0 ^∞ ((k+1−k)/(k(1+k)))dk =∫_0 ^∞ (dk/k)−∫_0 ^∞ (dk/(k+1)) =∣ln((k/(k+1)))∣_0 ^∞ some thing wrong

$$ \\ $$$$\sqrt{{tanx}}\:={tant} \\ $$$${tanx}={tan}^{\mathrm{2}} {t} \\ $$$${sec}^{\mathrm{2}} {xdx}=\mathrm{2}{tant}.{sec}^{\mathrm{2}} {tdt} \\ $$$${dx}=\frac{\mathrm{2}{tantsec}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{4}} {t}}{dt} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{tantsec}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{4}} {t}}×\frac{{t}}{{tan}^{\mathrm{2}} {t}}{dt} \\ $$$$\frac{{I}}{\mathrm{2}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {t}×\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {t}×\frac{{sint}}{{cost}}}×\frac{{cos}^{\mathrm{4}} {t}}{{sin}^{\mathrm{4}} {t}+{cos}^{\mathrm{4}} {t}}{dt} \\ $$$$\frac{{I}}{\mathrm{2}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\pi}{\mathrm{2}}−{t}\right)×\frac{\mathrm{1}}{{sintcost}}×\frac{{sin}^{\mathrm{4}} {t}}{{cos}^{\mathrm{4}} {t}+{sin}^{\mathrm{4}} {t}}{dt} \\ $$$${using}\:\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx} \\ $$$$\frac{{I}}{\mathrm{2}}+\frac{{I}}{\mathrm{2}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\pi}{\mathrm{2}}×\frac{{sin}^{\mathrm{4}} {t}}{{sintcost}}×\frac{\mathrm{1}}{{sin}^{\mathrm{4}} {t}+{cos}^{\mathrm{4}} {t}} \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{{tant}}×{sec}^{\mathrm{2}} {t}×\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{4}} {t}}×{dt} \\ $$$${a}={tant}\: \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\int_{\mathrm{0}} ^{\infty} \frac{{da}}{{a}\left(\mathrm{1}+{a}^{\mathrm{4}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{d}\left({a}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} \left(\mathrm{1}+{a}^{\mathrm{4}} \right)}\:\:\:\:{replace}\:{a}^{\mathrm{2}} \:{by}\:{b} \\ $$$$\frac{\mathrm{4}{I}}{\pi}=\int_{\mathrm{0}} ^{\infty} \frac{{db}}{{b}\left(\mathrm{1}+{b}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{bdb}}{{b}^{\mathrm{2}} \left(\mathrm{1}+{b}^{\mathrm{2}} \right)}\: \\ $$$$\frac{\mathrm{8}{I}}{\pi}=\int_{\mathrm{0}} ^{\infty} \frac{{dk}}{{k}\left(\mathrm{1}+{k}\right)}\:\:\:\left[{k}={b}^{\mathrm{2}} \right] \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{k}+\mathrm{1}−{k}}{{k}\left(\mathrm{1}+{k}\right)}{dk} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{dk}}{{k}}−\int_{\mathrm{0}} ^{\infty} \frac{{dk}}{{k}+\mathrm{1}} \\ $$$$=\mid{ln}\left(\frac{{k}}{{k}+\mathrm{1}}\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$\boldsymbol{{some}}\:\boldsymbol{{thing}}\:\boldsymbol{{wrong}} \\ $$$$ \\ $$

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