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

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

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

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

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

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

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

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

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

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

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